Answer:
a) 3
b) y=3(x-2)-2
Step-by-step explanation:
a)
Step 1: Find the gradient/slope of AB
[tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]=[tex]\frac{-2-2}{7-(-5)} =\frac{-4}{12} =-\frac{1}{3}[/tex]
Step 2: Find the gradient/slope of the line perpendicular to AB
The slope of the line perpendicular to AB is the reciprocal of the slope of the line AB multiplied by -1, so the slope of the line perpendicular to AB is 3.
b)
Step 1: Find the formula for the equation of a line
If you have a point and the slope, you can use the point-slope form, which is y-y1=m(x-x1).
Step 2: Plug in values to the formula
y-(-2)=3(x-2)
Step 3: Simplify
y+2=3(x-2)
y=3(x-2)-2
[tex]$a)$[/tex] The gradient of a line perpendicular to line [tex]$AB$[/tex] is [tex]$3$[/tex].
[tex]$b)$[/tex] The equation of the line perpendicular to line [tex]$AB$[/tex] and passing through point [tex]$C$[/tex] is [tex]y=-\frac{1}{3}x+\frac{13}{3}[/tex].
What is slope of perpendicular lines?
Lines that are perpendicular lines have slopes that are negative reciprocals.
It is given that the points,
[tex]$A=\left( -5,2 \right)$[/tex]
[tex]$B=\left( 7,-2 \right)$[/tex]
[tex]$C=\left( -2,5 \right)$[/tex]
[tex]$a)$[/tex] We will find the gradient of a line perpendicular to line [tex]$AB$[/tex] by using the slope formula.
So,
By the definition, the slope or gradient of a line describes its steepness, incline, or grade.
[tex]$m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$[/tex] which is equal to [tex]$\tan \left( \theta \right)$[/tex].
Where [tex]$m$[/tex] is the slope and [tex]$\theta $[/tex] is the angle of incline.
So,
Slope [tex]$\left( m \right)=\frac{-2-2}{7-\left( -5 \right)}$[/tex]
[tex]m=\frac{-1}{3}[/tex]
So, The slope of the line perpendicular to [tex]AB[/tex] is the reciprocal of the slope of the line [tex]AB[/tex] which is multiplied by [tex]-1[/tex].
so the slope of the line perpendicular to [tex]AB[/tex] is [tex]3[/tex].
[tex]b)[/tex] We will find the equation of the line perpendicular to [tex]AB[/tex] passing through the point [tex]C[/tex] that is [tex](-2,5)[/tex] by using the formula,
[tex]y=mx+b[/tex]
Here,
[tex]y=5[/tex]
[tex]x=-\frac{1}{3}[/tex]
We will find the [tex]y-[/tex] intercept of the line [tex](b)[/tex] first.
So,
[tex]$5=-\frac{1}{3}\left( -2 \right)+b$[/tex]
[tex]$5=\frac{2}{3}+b$[/tex]
Subtract [tex]\frac{2}{3}[/tex] both side.
So,
[tex]$b=\frac{13}{3}$[/tex]
So, the line perpendicular to [tex]AB[/tex] and passing throught the point [tex]C[/tex] is,
[tex]y=-\frac{1}{3}x+\frac{13}{3}[/tex].
Hence,
[tex]$a)$[/tex] The gradient of a line perpendicular to line [tex]$AB$[/tex] is [tex]$3$[/tex].
[tex]$b)$[/tex] The equation of the line perpendicular to line [tex]$AB$[/tex] and passing through point [tex]$C$[/tex] is [tex]y=-\frac{1}{3}x+\frac{13}{3}[/tex].
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