# Straight Line Questions

## Find the equation of line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0.

Solution : On solving the equations 4x + y – 1 = 0 and 7x – 3y – 35 = 0 by using point of intersection formula, we get x = 2 and y = -7 So, given lines intersect at (2, -7) Now, the equation of line joining the point (3, 5) and (2, …

## Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then

Solution : Since, ($$\alpha$$, -$$\alpha$$) lie on 4ax+2ay+c=0 and 5bx+2by+d=0. $$\therefore$$ 4a$$\alpha$$ + 2a$$\alpha$$ + c = 0 $$\implies$$ $$\alpha$$ = $$-c\over 2a$$  …..(i) Also, 5b$$\alpha$$ – 2b$$\alpha$$ + d = 0 $$\implies$$ $$\alpha$$ = $$-d\over 3b$$    …..(i) from equation (i) and (ii), $$-c\over 2a$$ = $$-d\over 3b$$ 3bc = 2ad Similar Questions Find …

## If PS is the median of the triangle, with vertices of P(2,2), Q(6,-1) and R(7,3), then equation of the line passing through (1,-1) and parallel to PS is

Solution : Since PS is the median, so S is the mid point of triangle PQR. So, Coordinates of S = ($${7+6\over 2}, {3 – 1\over 2}$$) = ($$13\over 2$$, 1) Slope of line PS = (1 – 2)/(13/2 – 2) = $$-2\over 9$$ Required equation passes through (1, -1) is y + 1 = …

## If $$\lambda x^2 – 10xy + 12y^2 + 5x – 16y – 3$$ = 0 represents a pair of straight lines, then $$\lambda$$ is equal to

Solution : Comparing with $$ax^2+2hxy+by^2+2gx+2fy+c$$ = 0 Here a = $$\lambda$$, b = 12, c = -3, f = -8, g = 5/2, h = -5 Using condition $$abc+2fgh-af^2-bg^2-ch^2$$ = 0, we have $$\lambda$$(12)(-3) + 2(-8)(5/2)(-5) – $$\lambda$$(64) – 12(25/4) + 3(25) = 0 $$\implies$$  -36$$\lambda$$ + 200 – 64$$\lambda$$ – 75 + 75 = …