The Three Lines 3x + 4y + 6 = 0,  and 4x + 7y + 8 = 0 Are

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Question

The three lines 3x + 4y + 6 = 0,  and 4x + 7y + 8 = 0 are

Solution

Correct option is

Concurrent

Solving the equations

            3x + 4y + 6 = 0, 4x + 7y + 8 = 0,

We get their point of intersection as (–2, 0). Since the coordinates of this point satisfy the equation,

            .

It following that the three lines are concurrent.

SIMILAR QUESTIONS

Q1

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

Q2

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

Q3

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

Q4

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 

Q5

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Q6

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Q7

Let P = (–1, 0), Q = (0, 0) and  be three points. Then the equation of the bisector of the angle PQR is

Q8

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Q9

The equation of the straight line which passes through the point (1, –2) and cuts off equal intercepts from the axes will be

Q10

The line (p + 2q)x + (p – 3q)y = p – q for different values of p and qpasses through the point