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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 = 0 Hence, 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)] = 0 Hence, 3cx^{2} – cy^{2} – 2xy + 2mx^{2} + 4y^{2} – 4mxy = 0 This implies (3c + 2m)x^{2} – 2(1+2m)xy + (4-c)y^{2} = 0 Now, the lines represented here are perpendicular to each other and hence, The coefficient of x^{2} + the coefficient of y^{2} = 0 This 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). Click here to download
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 = 0
Hence, 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)] = 0
Hence, 3cx^{2} – cy^{2} – 2xy + 2mx^{2} + 4y^{2} – 4mxy = 0
This implies (3c + 2m)x^{2} – 2(1+2m)xy + (4-c)y^{2} = 0
Now, the lines represented here are perpendicular to each other and hence,
The coefficient of x^{2} + the coefficient of y^{2} = 0
This 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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