{"id":20563,"date":"2022-08-12T12:07:21","date_gmt":"2022-08-12T06:37:21","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=20563"},"modified":"2023-03-30T14:29:25","modified_gmt":"2023-03-30T08:59:25","slug":"inter-1st-year-maths-1b-the-straight-line-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-1st-year-maths-1b-the-straight-line-important-questions\/","title":{"rendered":"Inter 1st Year Maths 1B The Straight Line Important Questions"},"content":{"rendered":"

Students get through Maths 1B Important Questions<\/a> Inter 1st Year Maths 1B The Straight Line Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 1st Year Maths 1B The Straight Line Important Questions<\/h2>\n

Question 1.
\nFind the equation of the straight line pass-ing through the point (2, 3) and making non-zero intercepts on the axes of co-ordinates whose sum is zero. [Mar 12]
\nSol. Equation of the line in the intercept form is
\n\\(\\frac{x}{a}\\) + \\(\\frac{y}{b}\\) = 1
\nGiven b = -a
\nEquation of the line is \\(\\frac{x}{a}\\) – \\(\\frac{y}{a}\\) = 1
\n\u21d2 x – y = a
\nThis line passes through (2, 3)
\n2 – 3 = a \u21d2 a = -1
\nEquation of the line is
\nx – y = -1 or x – y + 1 = 0<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the equation of the straight line passing through the points (at1<\/sub>2<\/sup>, 2at1<\/sub> ) and (at2<\/sub>2<\/sup>, 2at1<\/sub> ).
\nSolution:
\nEquation of the given points.
\n(x – x1<\/sub>) (y – y2<\/sub>) = (y – y1<\/sub>) (x1<\/sub> – x2<\/sub>)
\n( x -at1<\/sub>2<\/sup>) (2at1<\/sub> – 2at2<\/sub>)
\n= (y – 2at1<\/sub>) (at, \u2014 at2)
\n(x – at1<\/sub>2<\/sup>) .2a(t1<\/sub> – t2<\/sub>) = (y – 2at1<\/sub>) a. (t1<\/sub>2<\/sup> – t2<\/sub>2<\/sup>)
\n2x – 2at1<\/sub>2<\/sup> = y(t1<\/sub> + t2<\/sub>) – 2at1<\/sub>2<\/sup> + 2at1<\/sub>t2<\/sub> = 0
\n2x – (t1<\/sub> + t2<\/sub>)y + 2at1<\/sub>t2<\/sub> = 0<\/p>\n

Question 3.
\nFind the equation of the straight line passing through A (-1, 3) and i) parallel ii) perpendicular to the straight line passing through B(2, – 5) and C(4, 6). [Mar 11]
\nSolution:
\nSlope of BC = \\(\\frac{-5-6}{2-4}\\) = \\(\\frac{-11}{-2}\\) = \\(\\frac{11}{2}\\)
\ni) The required line is parallel to BC and passes through A(- 1, 3)
\nEquation of the parallel line is
\ny – 3 = \\(\\frac{11}{2}\\)(x + 1)
\n2y – 6 = 11x + 11
\n11x – 2y + 17 = 0<\/p>\n

ii) The required line is perpendicular to BC.
\nSlope of the required line = –\\(\\frac{1}{m}\\) = –\\(\\frac{2}{11}\\)
\nThis line passes through A (-1, 3)
\nEquation of the required line is
\ny – 3 = –\\(\\frac{2}{11}\\)(x + 1)
\n11y – 33 = -2x – 2
\n2x + 11y – 31 = 0<\/p>\n

\"Inter<\/p>\n

Question 4.
\nProve that the points (1, 11), (2, 15) and (-3, -5) are collinear and find the equation of the straight line containing them.
\nSolution:
\nA(1, 11), B(2, 15) and C (-3, -5) are the given points.
\nEquation of AB is
\n(y – y1<\/sub>) (x1<\/sub> – x2<\/sub>) = (x – x1<\/sub>) (y1<\/sub> – y2<\/sub>)
\n(y – 11) (1 -2) = (x – 1) (11 – 15)
\n-(y – 11) = -4 (x – 1)
\n-y + 11 = – 4x + 4
\n4x – y + 7 = 0
\nC (-3, -5) .
\n4x – y + 7 = 4(-3) + 5 + 7
\n= -12 + 12 = 0
\nC lines on AB \u21d2 A, B, C are collinear.
\nEquation of the line containing them is 4x – y + 7 = 0<\/p>\n

Question 5.
\nA straight line passing through A(1, -2) makes an angle tan-1<\/sup> \\(\\frac{4}{3}\\) with the positive direction of the X – axis in the anticlock-wise sense. Find the points on the straight line whose distance from A is 5.
\nSolution:
\n\"Inter
\nGiven \u03b1 = tan-1<\/sup> \\(\\frac{4}{3}\\) \u21d2 tan \u03b1 = \\(\\frac{4}{3}\\)
\n\"Inter
\ncos \u03b1 = \\(\\frac{3}{5}\\), sin \u03b1 = \\(\\frac{4}{5}\\)
\n(x1<\/sub>, y1<\/sub>) = (1, -2) = x1<\/sub> = 1, y1<\/sub> = -2
\nCase i) .
\nr = 5
\nx = x1<\/sub> + r cos \u03b1 = 1 + 5 . \\(\\frac{4}{5}\\) = 1 + 4 = 5
\ny = y1<\/sub> + r sin \u03b1= – 2 + 5 . \\(\\frac{3}{5}\\) = -2 + 3 = 1
\nCo-ordinates of B are (5, 1)<\/p>\n

Case ii) r = -5
\nx = x1<\/sub> + r cos \u03b1 = 1 – 5 . \\(\\frac{4}{5}\\) = 1 – 4 = -3
\ny = y1<\/sub> + r sin \u03b1 = -2 – 5 . \\(\\frac{3}{5}\\) = -2 – 3 = -5
\nCo-ordinates of C are (- 3, – 5)<\/p>\n

Question 6.
\nA straight line parallel to the line y = \\(\\sqrt{3} x\\) passes through Q(2, 3) and cuts the line 2x + 4y – 27 = 0 at P. Find the length of PQ.
\nSolution:
\nPQ is parallel to the straight line y = \\(\\sqrt{3} x\\)
\ntan \u03b1 = \\(\\sqrt{3}\\) = tan 60\u00b0
\n\u03b1 = 60\u00b0
\nQ(2, 3) is a given point
\n\"Inter
\nCo-ordinates of any point P are
\n(x1<\/sub> + r cos \u03b1, y1<\/sub> + r sin \u03b1)
\n(2 + r cos 60\u00b0, 3 + r sin 60\u00b0)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 7.
\nTransform the equation
\n3x + 4y+ 12 = 0 into
\ni) slope\u2014intercept form
\nii) Intercept form and
\nin) normal form
\nSolution:
\nThe given equation is 3x + 4y + 12=0
\ni) Slope-intercept form
\n4y = -3x – 12
\ny = (-\\(\\frac{3}{4}\\))x + (-3)
\nSlope = –\\(\\frac{3}{4}\\) , y – intercept = -3.<\/p>\n

ii) Intercept form
\n-3x – 4y = 12
\n–\\(\\frac{3 x}{12}\\) – \\(\\frac{4 y}{12}\\) = 1
\n\\(\\frac{x}{(-4)}\\) + \\(\\frac{y}{(-3)}\\) = 1
\nx – intercept = -4, y – intercept = -3<\/p>\n

iii) Normal form
\n-3x – 4y = 12
\nDividing with \\(\\sqrt{9+16}\\) = 5
\n(-\\(\\frac{3}{5}\\))x + (-\\(\\frac{4}{5}\\))y = \\(\\frac{12}{5}\\)
\nLet cos \u03b1 = \\(\\frac{-3}{5}\\) and sin \u03b1 = –\\(\\frac{4}{5}\\)
\np = \\(\\frac{12}{5}\\) so that
\nx cos \u03b1 + y sin \u03b1 = p
\n\u03b1 lies in third quadrant so that
\n\u03b1 = \u03c0 + tan-1<\/sup>(\\(\\frac{4}{3}\\))<\/p>\n

Question 8.
\nIf the area of the triangle formed by the straight line x = 0, y = 0 and 3x + 4y = a (a > 0) is 6, find the value of a. [May 11]
\nSolution:
\nEquation of the line is 3x + 4y = a
\n\"Inter
\nGiven \\(\\frac{a^{2}}{2 a}\\) = 6 \u21d2 a2<\/sup> = 144
\na = \u00b112
\nBut a > 0
\na = 12<\/p>\n

\"Inter<\/p>\n

Question 9.
\nFind the value of k, if the lines 2x – 3y + k = 0, 3x – 4y -13 = 0 and 8x – 11y – 33 = 0 are concurrent.
\nSolution:
\nLet L1<\/sub>, L2<\/sub>, L3<\/sub> be the straight lines whose equations are respectively
\n2x – 3y + k = 0 ………………. (1)
\n3x – 4y – 13 = 0 ……………….. (2)
\n8x – 11y – 33 = 0 ………………. (3)
\nSolving (2) and (3) for x and y
\n\"Inter
\nPoint of the lines (2) and (3) is (11, 5)
\nThe given lines L1<\/sub>, L2<\/sub>, L3<\/sub> are concurrent
\n\u2234 L1<\/sub> contain (11, 5)
\n\u2234 2(11) – 3(5) + k = 0
\nk = -7<\/p>\n

Question 10.
\nIf the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, then prove that a3<\/sup> + b3<\/sup> + c3<\/sup> = 3abc.
\nSolution:
\nThe equation of the given lines are
\nax + by +c = 0 ………………… (1)
\nbx + cy + a = 0 ……………….. (2)
\ncx + ay + b = 0 ……………….. (3)
\nSolving (1) and (2) point of intersection is got by
\n\"Inter
\nc(ab – c2<\/sup>) +a (bc – a2<\/sup>) + b(ca – b2<\/sup>) = 0
\nabc – c3<\/sup> + abc – a3<\/sup> + abc – b3<\/sup> = 0
\n\u2234 a3<\/sup> + b3<\/sup> + c3<\/sup> = 3abc.<\/p>\n

Question 11.
\nA variable straight line drawn through the point of intersection of the straight lines \\(\\frac{x}{a}\\) + \\(\\frac{y}{b}\\) = 1 and \\(\\frac{x}{b}\\) + \\(\\frac{y}{a}\\) = 1 meets the co-ordinate axes at A and B. Show that the locus of the mid point of \\(\\overline{\\mathrm{AB}}\\) is 2(a + b) xy = ab(x + y)
\nSolution:
\nEquations of the given lines are \\(\\frac{x}{a}\\) + \\(\\frac{y}{b}\\) = 1
\nand \\(\\frac{x}{b}\\) + \\(\\frac{y}{a}\\) = 1
\nSolving the point of intersection P(\\(\\frac{a b}{a+b}\\), \\(\\frac{a b}{a+b}\\))
\nQ (x0<\/sub>, y0<\/sub>) is any point on the locus
\n\u21d4 The line with x – intercept 2x0<\/sub>, y – intercept 2y0<\/sub>, passes through P
\n\u21d4 P lies on the straight line \\(\\frac{x}{2 x_{0}}\\) + \\(\\frac{y}{2 y_{0}}\\) = 1
\ni.e., \\(\\frac{a b}{a+b}\\)(\\(\\frac{1}{2 x_{0}}\\) + \\(\\frac{1}{2 y_{0}}\\)) = 1
\n\u21d2 \\(\\frac{a b}{a+b}\\) . \\(\\frac{x_{0}+y_{0}}{2 x_{0} y_{0}}\\) = 0
\nab(x0<\/sub> + y0<\/sub>) = 2(a + b) x0<\/sub>y0<\/sub>
\nQ(x0<\/sub>, y0<\/sub>) lies on the curve
\n2(a + b)xy = ab(x + y)
\nLocks the mid of point of
\nAB is 2(a + b)xy = ab(x +y)<\/p>\n

\"Inter<\/p>\n

Question 12.
\nIf a, b, c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.
\nSolution:
\na, b, c are in A.P.
\n2b = a + c
\na – 2b + c = 0
\na.1 + b(-2) + c = 0
\nEach number of family of straight lines
\nax + by + c = 0
\npasses through the fixed point (1,-2)
\n\u2234 Set of lines ax + by + c = 0 for parametric values of a, b and c is a family of concurrent lines.
\n\u2234 Point of concurrency is (1, -2).<\/p>\n

Question 13.
\nFind the value of k, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45\u00b0.
\nSolution:
\n\"Inter
\nSquaring and cross multiplying
\n2(4k + 5)2<\/sup> = 17(k2<\/sup> + 25)
\n2(16k2<\/sup> + 40k + 25) = 17k2<\/sup> + 425
\n32k2<\/sup> + 80k + 50 = 17k2<\/sup> + 425
\n15k2<\/sup> + 80k – 375 = 0
\n3k2<\/sup> + 16k – 75 = 0
\n(k – 3) (3k + 25) = 0
\nk = 3 or -25\/3<\/p>\n

Question 14.
\nFind the equation of the straight line passing through (x0<\/sub>, y0<\/sub>) and (i) parallel (ii) perpendicular to the straight line
\nax + by + c = 0.
\nSolution:
\nEquation of the given line is ax + by + c = 0
\ni) Equation of the parallel line is ax + by = k ………………… (1)
\nThis line passes through P(x0<\/sub>, y0<\/sub>) \u21d2 ax0<\/sub> + by0<\/sub> = k ……………… (2)
\nSubtracting (2) from (1) equation of the required line is a(x – x0<\/sub>) + b(y – y0<\/sub>) = 0<\/p>\n

ii) Equation of the perpendicular line is bx – ay = k.
\nThis line passes through P(x0<\/sub>, y0<\/sub>) \u21d2 bx0<\/sub> – ay0<\/sub> = k
\nSubtracting, equation of the required line is b(x – x0<\/sub>) – a(y – y0<\/sub>) = 0<\/p>\n

\"Inter<\/p>\n

Question 15.
\nFind the equation of the straight line perpendicular to the line 5x – 2y = 7 and passing through the point of intersection of the lines 2x + 3y = 1 and 3x + 4y = 6.
\nSolution:
\nGiven lines are L1<\/sub> = 2x + 3y – 1 = 0
\nL2<\/sub> = 3x + 4y – 6 = 0
\nEquation of the line passing through the intersection of L1<\/sub> = 0, L2<\/sub> = 0 is
\nL1<\/sub> + kL2<\/sub> = 0
\n(2x + 3y – 1) + k(3x + 4y – 6) = 0
\n(2 + 3k)x + (3 + 4k)y – (1 + 6k) = 0 …………………. (1)
\nThis line is perpendicular to 5x – 2y = 7 ………………… (2)
\na1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> = 0
\n5(2 + 3k) – 2(3 + 4k) = 0
\n10 + 15k – 6 – 8k = 0
\n7k = -4 \u21d2 k = – 4\/7
\nSubstituting in (1) equation of the required lines
\n(2 – \\(\\frac{12}{7}\\))x + (3 – \\(\\frac{16}{7}\\))x – (1 – \\(\\frac{24}{7}\\)) = 0
\n\\(\\frac{2}{7}\\)x + \\(\\frac{5}{7}\\)y + \\(\\frac{17}{7}\\) = 0 \u21d2 2x + 5y + 17 = 0<\/p>\n

Question 16.
\nIf 2x – 3y – 5 = 0 is the perpendicular bi-sector of the line segment joining (3, – 4) and (\u03b1, \u03b2) find \u03b1 + \u03b2.
\nSolution:
\n(\u03b1, \u03b2) is the reflection of (3, -4) is the line
\n2x – 3y – 5 = 0
\n\\(\\frac{\\alpha-3}{2}\\) = \\(\\frac{\\beta+4}{-3}\\) = \\(\\frac{-2(6+12-5)}{4+9}\\) = -2
\n\u03b1 – 3 = -4 \u21d2 \u03b1 = -1
\n\u03b2 + 4 = 6 \u21d2 \u03b2 = 2
\n\u03b1 + \u03b2 = -1 + 2 = 1<\/p>\n

\"Inter<\/p>\n

Question 17.
\nIf the four straight lines ax + by + p = 0, ax + by + q = 0, cx + dy + r = 0 and cx + dy + s = 0 form a parallelogram, show that the area of the parallelogram so formed is \\(\\left|\\frac{(p-a)(r-s)}{b c-a d}\\right|\\).
\nSolution:
\nLet L1<\/sub>, L2<\/sub>, L3<\/sub>, L4<\/sub> be the lines given by
\nL1<\/sub> = ax + by + p = 0
\nL2<\/sub> = ax + by + q = 0
\nL3<\/sub> = cx + dy + r = 0
\nL4<\/sub> = cx + dy + s = 0
\nL1<\/sub> and L2<\/sub> are parallel: L3<\/sub> and L4<\/sub> are parallel
\nArea of the parallelogram = \\(\\frac{\\mathrm{d}_{1} \\mathrm{~d}_{2}}{\\sin \\theta}\\)
\nd1<\/sub> = distance between L1<\/sub> and L2<\/sub> = \\(\\frac{|p-a|}{\\sqrt{a^{2}+b^{2}}}\\)
\nd2<\/sub> = distance between L3<\/sub> and L4<\/sub> = \\(\\frac{|r-s|}{\\sqrt{c^{2}+d^{2}}}\\)
\n\"Inter<\/p>\n

Question 18.
\nThe hypotenuse of a right angled isosceles triangle has its ends at the points (1, 3) and (-4, 1). Find the equations of the legs of the triangle.
\nSol:
\nLetA =(1, 3)and B = (-4, 1) and ABC b a right isosceles triangle with \\(\\stackrel{\\leftrightarrow}{A B}\\) as hypotenuse.
\nWe require, therefore, the equations of
\n\\(\\stackrel{\\leftrightarrow}{A C}\\) and \\(\\stackrel{\\leftrightarrow}{B C}\\)
\nSlope of \\(\\stackrel{\\leftrightarrow}{A B}\\) is \\(\\frac{1-3}{-4-1}\\) = \\(\\frac{2}{5}\\)
\nSince the slope of \\(\\stackrel{\\leftrightarrow}{A B}\\) is \\(\\), neither \\(\\frac{2}{5}\\) nor \\(\\stackrel{\\leftrightarrow}{A C}\\) is vertical.
\n\"Inter
\nTaking the slope of \\(\\stackrel{\\leftrightarrow}{A C}\\) as \\(\\frac{7}{3}\\), the slope of \\(\\stackrel{\\leftrightarrow}{B C}\\) would be –\\(\\frac{3}{7}\\). Therefore, the equations of \\(\\stackrel{\\leftrightarrow}{A C}\\) and \\(\\stackrel{\\leftrightarrow}{B C}\\) are respectively.
\ny – 3 = \\(\\frac{7}{3}\\) (x – 1) and y – 1 = – \\(\\frac{3}{7}\\) (x + 4),
\nwhich become 7x – 3y + 2 = 0 and 3x + 7y + 5 = 0
\nIf the lines drawn through A and B respectively parallel to \\(\\stackrel{\\leftrightarrow}{B C}\\) and \\(\\stackrel{\\leftrightarrow}{A C}\\) meet at D, then \u2206ABD is also right isosceles, having \\(\\stackrel{\\leftrightarrow}{A B}\\) as its hypotenuse.
\nTherefore, the equations of \\(\\stackrel{\\leftrightarrow}{A D}\\) and \\(\\stackrel{\\leftrightarrow}{B D}\\) are respectively,
\n3(x – 1) + 7(y – 3) = 0 and 7(x + 4) – 3(y – 1) = 0
\n\u21d2 3x + 7y – 24 = 0 and 7x – 3y + 31 = 0.
\nTherefore, the two pairs of legs required are
\n7x – 3y + 2 = 0, 3x + 7y + 5 = 0 and
\n3x + 7y – 24 = 0, 7x – 3y + 31 = 0.
\nNote : ADBC is square.<\/p>\n

\"Inter<\/p>\n

Question 19.
\nA line is such that its segment between the lines 5x – y + 4 = 0 and 3x + 4y – 4 = 0 is . bisected at the point (1, 5). Obtain its equation.
\nSolution:
\nLet the required line meet 3x + 4y – 4 = 0 at A and 5x – y + 4 = 0 at B, so that AB is the segment between the given lines, with its mid-point at C = (1, 5).
\nThe equation 5x – y + 4 = 0 can be written as
\ny = 5x + 4 so that any point on \\(\\stackrel{\\leftrightarrow}{B X}\\) is (t, 5t + 4) for all real t.
\n\u2234 B = (t, 5t + 4) for some t. Since (1, 5) is the mid-point of \\(\\stackrel{\\leftrightarrow}{A B}\\)
\nA = [2 – t, 10 – (5t + 4)]
\n= [2 – t, 6 – 5t]
\nSince A lies on 3x + 4y – 4 = 0,
\n3(2 – 1) + 4(6 – 5t) – 4 = 0
\n\u21d2 -23t + 26 = 0
\n\u21d2 t = \\(\\frac{26}{23}\\)
\n\"Inter
\nequation of \\(\\stackrel{\\leftrightarrow}{A B}\\) is y – 5 = \\(\\frac{107}{3}\\) (x – 1)
\n\u21d2 3y – 15 = 107x – 107
\n\u21d2 107x – 3y – 92 = 0.<\/p>\n

Question 20.
\nAn equilateral triangle has its incentre at the origin and one side as x + y – 2 = 0. Find the vertex opposite to x + y – 2 = 0.
\nSolution:
\nLet ABC be the equilateral triangle and
\nx + y – 2 = 0 represent side \\(\\stackrel{\\leftrightarrow}{B C}\\).
\nSince O is the incentre of the triangle, \\(\\stackrel{\\leftrightarrow}{A D}\\) is the bisector of \u2220BAC .
\nSince the triangle is equilateral, \\(\\stackrel{\\leftrightarrow}{A D}\\) is the perpendicular bisector of \\(\\stackrel{\\leftrightarrow}{B C}\\).
\nSince O is also the centroid, AO : OD = 2 : 1. [The centroid, circumcentre incentre and orthocentre coincide]
\nLet D = (h, k).
\nSince D is the foot of the perpendicular from O onto \\(\\stackrel{\\leftrightarrow}{B C}\\), D is given by
\n\"Inter
\n\u2234 x1<\/sub> = -2, y1<\/sub> = -2.
\n\u2234 A = (-2, -2), the required vertex.<\/p>\n

\"Inter<\/p>\n

Question 21.
\nFind the orthocentre of the triangle whose vertices are (-5,-7) (13, 2) and (-5, 6)
\nSolution:
\nLet A(-5, -7), B(13,2) and C (-5, 6) be the vertices of a triangle. Let \\(\\stackrel{\\leftrightarrow}{A D}\\) be the perpendicular drawn from A to \\(\\stackrel{\\leftrightarrow}{B C}\\) and \\(\\stackrel{\\leftrightarrow}{B E}\\) be the perpendicular drawn from B to \\(\\stackrel{\\leftrightarrow}{A C}\\).
\n\"Inter
\nNow slope of \\(\\stackrel{\\leftrightarrow}{B C}\\) = \\(\\frac{6-2}{-5-13}\\) = \\(\\frac{-2}{9}\\)
\nSince \\(\\stackrel{\\leftrightarrow}{A D}\\) \u22a5 \\(\\stackrel{\\leftrightarrow}{B C}\\) , slope of \\(\\stackrel{\\leftrightarrow}{A D}\\) = \\(\\frac{9}{2}\\) and so, the equation of \\(\\stackrel{\\leftrightarrow}{A D}\\) is
\n9x – 2y = – 45 + 14 = -31 ………………… (1)
\nEquation of \\(\\stackrel{\\leftrightarrow}{A C}\\) is x = -5 which is a vertical line and therefore equation of \\(\\stackrel{\\leftrightarrow}{B E}\\) is y = 2. ……………… (2)
\nPoint of intersection of the lines (1) and (2) is (-3, 2) which is the orthocentre of \u2206 ABC.<\/p>\n

Question 22.
\nIf the equations of the sides of a triangle are 7x + y – 10 = 0, x – 2y + 5 = 0 and x + y + 2 = 0, find the orthocentre of the triangle.
\nSolution:
\nLet \u2206 ABC be the given triangle
\nLet the equations x – 2y + 5 = 0 ……………… (1)
\n7x + y – 10 = 0 ……………… (2)
\nand x + y + 2 = 0 ……………….. (3)
\nrepresent the sides \\(\\overleftrightarrow{\\mathrm{AB}}\\), \\(\\overleftrightarrow{\\mathrm{BC}}\\), and \\(\\overleftrightarrow{\\mathrm{CA}}\\)
\n\"Inter
\nLet \\(\\overleftrightarrow{\\mathrm{AD}}\\) and \\(\\overleftrightarrow{\\mathrm{BE}}\\) be the altitudes drawn from A and B respectively to the sides \\(\\overleftrightarrow{\\mathrm{BC}}\\) & \\(\\overleftrightarrow{\\mathrm{CA}}\\)
\nSolving the equations (1) and (3), we obtain A =(-3,1).
\nSince \\(\\overleftrightarrow{\\mathrm{AD}}\\) \u22a5 \\(\\overleftrightarrow{\\mathrm{BC}}\\), the equations of \\(\\overleftrightarrow{\\mathrm{AD}}\\) is
\nx – 7y = -3 – 7 = -10 …………………… (4)
\nSolving the equation (1) and (2), we obtain B = (1, 3).
\nSince \\(\\overleftrightarrow{\\mathrm{BE}}\\) \u22a5 \\(\\overleftrightarrow{\\mathrm{AC}}\\) , the equation of BE is
\nx – y = 1 – 3 = -2 …………… (5)
\nPoint of intersection of the lines (4) and (5) is H(\\(\\frac{-2}{3}\\), \\(\\frac{4}{3}\\)) which is the orthocentre of \u2206 ABC.<\/p>\n

\"Inter<\/p>\n

Question 23.
\nFind the circumcentre of the triangle whose vertices are (1, 3), (-3, 5) and (5, -1).
\nSolution:
\nLet the vertices of the triangle be
\nA(1, 3), B(-3, 5) and (5,-1).
\nThe midpoints of the sides \\(\\overline{\\mathrm{BC}}\\), \\(\\overline{\\mathrm{CA}}\\) are respectively D(1, 2) and E(3, 1).
\nLet S be the point of intersection of the perpendicular bisectors of the sides \\overline{\\mathrm{BC}} and \\overline{\\mathrm{CA}}
\n\"Inter
\nSince the slope of \\(\\overleftrightarrow{\\mathrm{BC}}\\) = \\(\\frac{5+1}{-3-5}\\) = \\(\\frac{-3}{4}\\), the slope \\(\\overleftrightarrow{\\mathrm{SD}}\\) is \\(\\frac{4}{3}\\) and therefore its equation is 4x – 3y = 4 – 6 = -2 …………….. (1)
\nSlope of \\(\\overleftrightarrow{\\mathrm{AC}}\\) = \\(\\frac{3+1}{1-5}\\) = -1 \u21d2 Slope of –\\(\\sqrt{3x}\\) = 1
\n\u2234 Equation of \\(\\overleftrightarrow{\\mathrm{SE}}\\) is x – y = 3 – 1 = 2 ………………….. (2),
\nSolving the equations (1) and (2),we obtain S = (-8, -10) which is the circumcentre of \u2206 ABC.<\/p>\n

Question 24.
\nFind the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0. [May 11, 05; Mar. 06]
\nSolution:
\nLet the given equations 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0 represent the sides \\(\\overleftrightarrow{\\mathrm{BC}}\\), \\(\\overleftrightarrow{\\mathrm{CA}}\\) and \\(\\overleftrightarrow{\\mathrm{AB}}\\) respectively of \u2206 ABC.
\nSolving the above equations two by two, we obtain the vertices A(-2, 3), B(1, -2) and (2, 1) of the given triangle.
\nThe midpoints of the sides \\(\\overline{\\mathrm{BC}}\\) and \\(\\overline{\\mathrm{CA}}\\) are respectively D = (\\(\\frac{3}{2}\\), \\(\\frac{-1}{2}\\)) and E = (0, 2).
\n\"Inter
\nEquation of \\(\\overleftrightarrow{\\mathrm{SD}}\\) the perpendicular bisector of \\(\\overline{\\mathrm{BC}}\\) is x + 3y = 0 and the equation of \\(\\overleftrightarrow{\\mathrm{SE}}\\) the perpendicular bisector of \\(\\overline{\\mathrm{AC}}\\) is 2x – y + 2 = 0. Solving these two equations, we obtain the point of intersection of the lines \\(\\overleftrightarrow{\\mathrm{SD}}\\) and \\(\\overleftrightarrow{\\mathrm{SE}}\\) which is S(\\(\\frac{-6}{7}\\), \\(\\frac{2}{7}\\)) the circumcentre of \u2206 ABC.<\/p>\n

\"Inter<\/p>\n

Question 25.
\nFind the incentre of the triangle formed by the straight lines y = \\(\\sqrt{3}\\)x, y = –\\(\\sqrt{3}\\)x and y = 3.
\nSolution:
\n\"Inter
\nThe straight lines y = \\(\\sqrt{3x}\\) and y = –\\(\\sqrt{3x}\\) respectively make angles 60\u00b0 and 120\u00b0 with the positive direction of X – axis.
\nSince y = 3 is a horizontal line, the triangle formed by the three given lines is equilateral. So in-centre is same and centroid.
\nVertices of the triangle and O(0, 0), A(\\(\\sqrt{3}\\), 3) and D(\\(\\sqrt{3}\\), 3).
\n\u2234 Incentre is (\\(\\frac{0+\\sqrt{3}-\\sqrt{3}}{3}\\), \\(\\frac{0+3+3}{3}\\)) = (0, 2)<\/p>\n

Question 26.
\nFind the equation of the straight line whose distance from the origin is 4, If the normal ray from the origin to the straight line makes an angle of 135\u00b0 with the positive direction of the x-axis.
\nSolution:
\nThe equation of the given line is
\nx cos \u03b1 + y sin \u03b1 = p where p = 4 and \u03b1 = 135\u00b0
\n\u2234 x(\\(\\frac{-1}{\\sqrt{2}}\\)) + y(\\(\\frac{1}{\\sqrt{2}}\\)) = 4 \u21d2 or x – y + 4\\(\\sqrt{2}\\) = 0<\/p>\n

(i). Transform the equation x + y + 1 = 0 in to normal from.
\nSolution:
\nx + y + 1 = 0
\n\u21d4 (\\(\\frac{-1}{\\sqrt{2}}\\))n + (\\(\\frac{-1}{\\sqrt{2}}\\)) y = \\(\\frac{1}{\\sqrt{2}}\\)
\n\u21d4 x cos \\(\\frac{5\\pi}{42}\\) + y sin \\(\\frac{5\\pi}{4}\\) = \\(\\frac{1}{\\sqrt{2}}\\)
\nHence the normal form of the equation of the given straight line is x cos \\(\\frac{5\\pi}{4}\\) + y sin \\(\\frac{5\\pi}{4}\\) = \\(\\frac{1}{\\sqrt{2}}\\) and the distance of the line from the origin is = \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

ii) A straight line passing through A(1, -2) makes and angle Tan-1<\/sup>\\(\\frac{4}{3}\\) with the positive direction of the x – axis in the anti-clock wise sense. Find the points of the straight line whose distance from A is 5.
\nSolution:
\nThe paramatic equation of the lines A (1, -2)
\nSlope is \\(\\frac{4}{3}\\) (\u2235 tan \u03b8 = \\(\\frac{4}{3}\\))
\nare x = 1 + r cos \u03b8 = 1 + r(\\(\\frac{3}{5}\\)) and y
\n= -2 + r sin\u03b8
\n= -2 + r (\\(\\frac{4}{5}\\))
\nThe points on the above line a + a distance of (r) = 5 correspond to r = \u00b15 in the above equation and are
\n\u2234 (4, 2) and (-2, -6)<\/p>\n

\"Inter<\/p>\n

Question 27.
\nTrans form the equation 3x + 4y + 12 = 0 in to (i) slope – intercept from (ii) intercept form (iii) normal form.
\nSolution:
\n(i) slop – intercept form
\n3x + 4y + 12 = 0
\n\u21d4 4y = -3x – 12
\n\u21d4 y = (\\(\\frac{-3}{4}\\)) x + (-3)
\n\u2234 Slope = (\\(\\frac{-3}{4}\\)) and y – intercept = -3,<\/p>\n

ii) Intercept form :
\n3x = 4y + 12 = 0
\n\u21d4 \\(\\frac{-3y}{12}\\) – \\(\\frac{-4y}{12}\\) = 1
\n\u21d4 \\(\\frac{x}{(-4)}\\) + \\(\\frac{y}{(-3)}\\) = 1
\n\u2234 x-intercept of the line is -4, and tan y-intercept is -3.<\/p>\n

iii) Normal form
\n3x = 4y + 12 = 0
\n\u21d4 -3x + -4y – 12 = 0
\n\u21d4 (\\(\\frac{-3}{5}\\))x + (\\(\\frac{-4}{5}\\))y = \\(\\frac{12}{5}\\)
\n\u21d4 x cos \u03b1 + y sin \u03b1 = p \u21d2 p\\(\\frac{12}{5}\\) and cos \u03b1 = \\(\\frac{-3}{5}\\), sin \u03b1 = \\(\\frac{-4}{5}\\) determine the angle \u03b1 in (0, 2\u03c0).<\/p>\n

Question 28.
\nFind the angle between the lines 2x + y + 4 = 0 and y – 3x = 7.
\nSolution:
\nThe angle between the given lines
\n= cos-1<\/sup> \\(\\frac{-6+1}{\\sqrt{5 \\times 10}}\\)
\n= cos-1<\/sup> [latex]\\frac{5}{\\sqrt{2}}[\/latex] = cos-1<\/sup>(\\(\\frac{1}{\\sqrt{2}}\\)) = \\(\\frac{\\pi}{4}\\)<\/p>\n

\"Inter<\/p>\n

Question 29.
\nFind Q(h,k) in the foot of the perpendicular from p(x1<\/sub>, y1<\/sub>) on the straight lines ax + by + c = 0 then (h – x1<\/sub>) ; a = (k – y1<\/sub>); b = -(ax1<\/sub> + by1<\/sub> + c); (a2<\/sup> + b2<\/sup>)
\nSolution:
\nEquation of \\(\\stackrel{\\leftrightarrow}{P Q}\\) which is normal to the given straight line
\nL : ax + by + c = 0
\nbx – ay = bx1<\/sub> – ay1<\/sub>
\n\"Inter
\n\u2234 Q \u2208 \\(\\stackrel{\\leftrightarrow}{P Q}\\) we have
\nbh – ak = bx1<\/sub> – ay1<\/sub>
\n\u2234 b(h – x1<\/sub>) = a(k – y1<\/sub>)
\nor (h – x1<\/sub>) a = (k – y1<\/sub>); b.
\nBut, this implus that h = a\u03bb + x1<\/sub>, and k = b\u03bb + y1<\/sub>
\nfor some \u03bb \u2208 R. sine Q(h, k) in point on L.
\na(a\u03bb + x1<\/sub>) + b(b\u03bb + y1<\/sub>) + c = 0
\ni.e., \u03bb = \\(\\frac{\\left(a x_{1}+b y_{1}+c\\right.}{\\left(a^{2}+b^{2}\\right)}\\)
\n\u2234 (h – x1<\/sub>) ; a = (k – y1<\/sub>); b
\n= – (ax1<\/sub> + by1<\/sub> + c); (a2<\/sup> + b2<\/sup>)<\/p>\n

Question 30.
\nFind the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y -1 = 0.
\nSolution:
\nThe equation of the given straight lines is 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0.
\nformula : \\(\\frac{c_{1}-c_{2}}{\\sqrt{a^{2}+b^{2}}}\\) = \\(\\frac{-6+1}{\\sqrt{6^{2}+8^{2}}}\\) = \\(\\frac{5}{10}\\) = \\(\\frac{1}{2}\\)<\/p>\n

Question 31.
\nFind the condition for the points (a, o) (h, k) and (o, b) when ab \u2260 0 to be collinear. [Mar 10]
\nSolution:
\nThe equation of the line passing through (a, o) and (o, b) is \\(\\frac{x}{a}\\) + \\(\\frac{y}{b}\\) = 1
\nThe given points are collinear \u21d2 (h, k) lies on the above line \u21d2 \\(\\frac{h}{a}\\) + \\(\\frac{k}{b}\\) = 1
\n\u21d2 hb + ka = ab<\/p>\n

\"Inter<\/p>\n

Question 32.
\nFind the area of the triangle formed by the straight lines x cos a + y sin a = p and the axes of co-ordinates. [Mar 10]
\nSolution:
\nThe area of the triangle formed by the line ax + by + c = 0
\nand the co-ordinate axes is \\(\\frac{c^{2}}{2|a b|}\\)
\n\u2234 Area of the triangle = \\(\\frac{p^{2}}{2|\\cos \\alpha \\cdot \\sin \\alpha|}\\)
\n= \\(\\frac{p^{2}}{|\\sin 2 \\alpha|}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 1B Important Questions Inter 1st Year Maths 1B The Straight Line Important Questions which are most likely to be asked in the exam. Intermediate 1st Year Maths 1B The Straight Line Important Questions Question 1. Find the equation of the straight line pass-ing through the point (2, 3) and making non-zero … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":39435,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[13],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/The-Straight-Line-Important-Questions.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/20563"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=20563"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/20563\/revisions"}],"predecessor-version":[{"id":20711,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/20563\/revisions\/20711"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/39435"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=20563"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=20563"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=20563"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}