Question

Given two intersecting straight lines inclined at an angle a, intersecting at the origin, MA and MB are perpendicular from variable point M (2t –1, t + 2) on the two given intersecting straight lines meeting at N, then the locus of N is a –

• A. Point
• B. Straight line
• C. Pair of straight line
• D. Circle

Answer 1

Answer: (B)

Straight line

Answer 2

Lines are x = 0, y = mx

A (2t –1, 0)

solving the lines y = mx and

(y – (t–2)) = – $\frac{1}{m}$ (x – (2t –1)) we get

$\left( \frac{2t - 1 + m(t + 2)}{1 + m^{2}},\frac{2tm - m + m^{2}(t + 2)}{1 + m^{2}} \right)$

Let N(h, k) lies on BN, AN

equation of line BN є k = $\frac{2tm - m + m^{2}(t + 2)}{1 + m^{2}}$

AN є k = m(h – (2t –1))

using value of t, t = $\frac{mh + m - k}{2m}$ in BN

2mh – (4 + 2m² + m) k + 5m³ = 0

which is straight line.