Question

The three lines 4x – 7y + 10 = 0, x + y = 5 and 7x + 4y = 15 from the sides of a triangle. The line (1, 2) is its

Solution

Correct option is

Orthocenter

(1, 2) is the intersection of two perpendicular lines 4x – 7y + 10 = 0 and 7x + 4y = 15. It is therefore its orthocenter.  

SIMILAR QUESTIONS

Q1

Let PQR be a right angled isosceles triangle, right angled at P(2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is

Q2

The point (2, 1) is shifted through a distance  units measured parallel to the line x + y = 1 in the direction of decreasing ordinates to reach Q.The image of Q w.r.t. given line is

Q3

Given the family of lines a(2x + y + 4) + b(x – 2y – 3) = 0. The number of lines belonging to the family at a distance  from any point (2, –3) is

Q4

 

Given four lines with equations x + 2y – 3 = 0, 3x + 4y – 7 = 0,

2x + 3y – 4 = 0, 4x + 5y – 6 = 0, then

Q5

Area of the parallelogram formed by the lines y = mx, y = mx + 1, y = nxand y = nx + 1 equals.

Q6

The area bounded by the curves 

Q7

A line making an angle  with the + ive direction of x-axis passes throughP(5, 6) to meet the line x = 6 at Q and y = 9 at R the QR is

Q8

A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. Then the point Odivides the segment PQ in the ratio

Q9

Orthocenter of triangle whose vertices are (0, 0), (3, 4), (4, 0) is

Q10

The equation to the side of a triangle are x – 3y = 0, 4x + 3= 5 and 3x + y = 0. The line 3x – 4y = 0 passes through