Let's solve this step-by-step:
### Given Points:
- Point [tex]\( A = (-5, 2) \)[/tex]
- Point [tex]\( B = (7, -2) \)[/tex]
- Point [tex]\( C = (-2, 5) \)[/tex]
### Part (a): Find the Gradient of the Line Perpendicular to Line AB
1. Find the Gradient (Slope) of Line AB:
- The formula to calculate the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
\text{Slope}_{AB} = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
- Substituting the coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[
\text{Slope}_{AB} = \frac{-2 - 2}{7 - (-5)} = \frac{-4}{12} = -\frac{1}{3}
\][/tex]
2. Find the Gradient of the Perpendicular Line:
- The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope.
[tex]\[
\text{Slope}_{\text{perpendicular}} = -\frac{1}{\left(-\frac{1}{3}\right)} = 3
\][/tex]
- Thus, the gradient of the line perpendicular to line [tex]\( AB \)[/tex] is:
[tex]\[
\boxed{3}
\][/tex]
### Part (b): Find the Equation of the Line Perpendicular to Line AB and Passing Through Point C
1. Use the Point-Slope Form of the Equation of a Line:
- The equation of a line in point-slope form is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\( m \)[/tex] is the slope.
2. Substitute the Point [tex]\( C \)[/tex] and the Slope:
- Point [tex]\( C = (-2, 5) \)[/tex] and the perpendicular gradient [tex]\( m = 3 \)[/tex]:
[tex]\[
y - 5 = 3(x - (-2))
\][/tex]
- Simplify and solve for [tex]\( y \)[/tex]:
[tex]\[
y - 5 = 3(x + 2)
\][/tex]
[tex]\[
y - 5 = 3x + 6
\][/tex]
[tex]\[
y = 3x + 6 + 5
\][/tex]
[tex]\[
y = 3x + 11
\][/tex]
- Thus, the equation of the line is:
[tex]\[
\boxed{y = 3x + 11}
\][/tex]
To summarize:
- The gradient of the line perpendicular to line [tex]\( AB \)[/tex] is [tex]\( 3 \)[/tex].
- The equation of the line perpendicular to line [tex]\( AB \)[/tex] and passing through point [tex]\( C \)[/tex] is [tex]\( y = 3x + 11 \)[/tex].
