Solved Examples on Straight Lines

Illustration 1:The vertices of a triangle are A (-1, -7), B (5, 1) and C (1, 4). Then what is the equation of the bisector of the angle ∠ABC?(1993)

Solution:The vertices of the triangle are given to be A (-1, -7), B (5, 1) and C (1, 4).Hence, the equation of the line AB is (y-1) = [1-(-7)] / [5-(-1)] (x-5)

Hence, (y-1) = 8/6 (x-5)

This gives (y-1) = 4/3 (x-5)

We have 3y - 3 = 4 x – 20.

Hence, the equation becomes 3y-4x+17 = 0

Equation of the line BC is (y-4) = (4-1)/(1-5) (x-1)

Hence, (y-4) = -3/4(x-1)

This yields, 4y-16 = -3x+3

Hence, 3x+4y-19 = 0

Again, the equation of the bisectors of angles between two given lines AB and BC are

(3y-4x+17)/√3

^{2}+4^{2}= ± (4y+3x-19) / √4^{2}+ 3^{2}This gives (3y-4x+17) = ± (4y+3x-19)

Hence, we have two relations

(3y-4x+17) = (4y+3x-19)

and (3y-4x+17) = - (4y+3x-19)

This gives y +7x = 36 and 7y-x = 2

Out of these two, the equation of the bisector of ∠ABC is 7y = x + 2.

Illustration 2:A line cuts the x-axis at A (7, 0) and the y-axis at B (0, -5). A variable line PQ is drawn perpendicular to AB cutting the x-axis in P and the y-axis in Q. If AQ and BP intersect at R, then find the locus of R.(1990)

Solution:The given line cuts the x-axis at A (7, 0) and the y-axis at B (0, -5).Hence, the equation of the line AB is x/7 + y/-5 = 1 …… (1)

Hence, the equation becomes 5x-7y = 35.

We know that the equation of the line perpendicular to AB is

7x+5y = μ.

Hence, it will meet x-axis at P (μ/7, 0) and y-axis at Q (0, μ/5).

The equations of lines AQ and BP are x/7 + 5y/μ = 1 and 7x/μ – y/5 =1 respectively.

Let R(h, k) be the point of intersection of lines AQ and BP.

Then, h/7 + 5k/μ = 1

And 7h/μ – k/5 = 1

This gives 1/5k (1-h/7) = 1/7h (1+k/5)

So, h(7-h) = k(5+k)

This gives h

^{2}+ k^{2}– 7h + 5k = 0Hence, the locus of the point is x

^{2}+y^{2}-7x+5y = 0.

Illustration 3:Show that all chords of curve 3x^{2}– y^{2}– 2x + 4y = 0, which subtend a right angle at the origin pass through a fixed point. Find the coordinates of that point.(1991)

Solution:The given curve is 3x^{2}– y^{2}– 2x + 4y = 0. ….. (1)Let y = mx be the chord of the curve (1) which subtends a right angle at the origin. Then, we try to find the points of intersection of the given curve with the chord y = mx + c.

Hence, we have 3x

^{2}– y^{2}– 2x [(y-mx)/c)] + 4y [(y-mx)/c)] = 0Hence, 3cx

^{2}– cy^{2}– 2xy + 2mx^{2}+ 4y^{2}– 4mxy = 0This implies (3c + 2m)x

^{2}– 2(1+2m)xy + (4-c)y^{2}= 0Now, the lines represented here are perpendicular to each other and hence,

The coefficient of x

^{2}+ the coefficient of y^{2}= 0This gives, 3c + 2m + 4 – c = 0

Hence, c+m+2 = 0

On comparing with y = mx + c we get that y = mx + c passes through (1, -2).

Hence, the coordinates of the required point are (1, -2).

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