Question

What is an equation of a line which passes through (6,9) and is perpendicular to the line whose equation is 4x − 6y = 15?
3
1) y−9=−2(x−6)
2
2) y−9= 3(x−6)
3
3) y+9=−2(x+6)
4) y+9=23(x+6)

Answer 1

Given:

The equation of the line passes through the point (6,9) and is perpendicular to the line whose equation is $4 x-6 y=15$

We need to determine the equation of the line.

Slope:

Let us convert the equation to slope - intercept form.

$-6 y=15-4x$

   $y=\frac{2}{3}x-\frac{5}{2}$

From the above equation, the slope is $m_1=\frac{2}{3}$

Since, the lines  are perpendicular, the slope of the line can be determined using the formula,

$m_1 \cdot m_2=-1$

  $\frac{2}{3} \cdot m_2=-1$

      $m_2=-\frac{3}{2}$

Therefore, the slope of the equation is $m=-\frac{3}{2}$

Equation of the line:

The equation of the line can be determined using the formula,

$y-y_1=m(x-x_1)$

Substituting the point (6,9) and the slope $m=-\frac{3}{2}$ in the above formula, we get;

$y-9=-\frac{3}{2}(x-6)$

Simplifying the terms, we get;

$2(y-9)=-3(x-6)$

$2y-18=-3x+18$

$3x+2y=36$

Thus, the equation of the line is $3x+2y=36$