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The lines 2y+3x16=0 and 7y2x6=0 inter...

The lines 2y+3x16=0 and 7y2x6=0 intersect at point P. Find the coordinates of P

A.

(4, 2)

B.

(4, -2)

C.

(-4, 2)

D.

(-4, -2)

Correct answer is A

The point of intersection for the two lines exists at the point where the two lines are equal to each other. Make anyone of the variables the subject of the formula and equate the two lines to each other and solve for the coordinates of the point of intersection.

Given lines 2y+3x16=0 and 7y2x6=0 , making x the subject of the formula:

Line 1 : 2y+3x16=03x=162y

Line 2 : 7y - 2x - 6 = 0 \implies -2x = 6 - 7y

\therefore x = \frac{7}{2}y - 3

Equating them together and solving, we have:

\frac{16}{3} - \frac{2}{3}y = \frac{7}{2}y - 3  \implies \frac{16}{3} + 3 = \frac{7}{2}y + \frac{2}{3}y

\frac{25}{3} = \frac{25}{6}y \therefore y = 2

Putting y = 2 in the equation 3x = 16 - 2y, we have

3x = 16 - 2(2) = 16 - 4 = 12 \implies x = 4

The coordinate of P is (4, 2).