What is the equation of the straight line passing through (4,1) which is parallel to the line given by the equation 3x+2y=12?

Question

• A. 3y=14-2x
• B. 2y=11-3x
• C. 2y=14-3x
• D. 3y=2x-5
• E. 2y=3x-10

Answer 1

3x+2y=12 2y= - 3x+12 y= - 3/2*x +12 m = - 3/2 and (4, 1) m= - 3/2 and x(1) =4 and y(1) = 1 y-y(1) = m(x - x(1)) y - 1 = - 3/2( x - 4) y = - 3/2 * x + 3/2 * 4 + 1 y = - 3/2 * x + 7 2y = - 3x + 14 The answer is C

Answer 2

For parallel lines, they have equivalent gradients; thus we rewrite equation above in the form y=mx+c; where m is the gradient and c is y-intercept. => y=-3/2x+4; we have point(4,1) and we introduce another point say (x,y) on the second line. equation of line two will be (y-1)/(x-4)=-3/2; 2y-2=-3x+12 collect like terms together, 2y=–3x+14, thus answer is 2y=–3x+14