Question

Find the coordinates of the foot of perpendicular, image and equation of perpendicular drawn from the (2, 3) to the line y=3x – 4.

Answer 1

- Hint: Take the point of the image as (h,k) then find the midpoint of (h,k) and (2,3) which lies on equation y = 3x – 4. Then substitute this to find the relation between h and k. Now as the line joining point and image is perpendicular to the line, find another relation between h and k using slope relation. Then find the values of (h,k), the foot of perpendicular and finally the equation.

Complete step-by-step solution -

In the question we have to find the foot of perpendicular, image and equation of perpendicular drawn from (2,3) to the line y = 3x – 4.
So we can represent it as.

Let the image be represented as (h, k) about the line y = 2x -3 of (2, 3).
Now we know the formula, if points are $(x_1,y_1),(x_2,y_2)$ and its midpoint is (x, y), then
$x={\displaystyle \frac{x_1+x_2}{2}},y={\displaystyle \frac{y_1+y_2}{2}}$ .
So, the midpoint of (h,k) and (2,3) is equal to $({\displaystyle \frac{h+2}{2}},{\displaystyle \frac{k+3}{2}})$ .
We know that $({\displaystyle \frac{h+2}{2}},{\displaystyle \frac{k+3}{2}})$ lies on the line y = 3x – 4, then we can say that the point also satisfies the line.
Hence we can write it as,
${\displaystyle \frac{k+3}{2}}=3\left({\displaystyle \frac{h+2}{2}}\right)-4$
Now further simplifying we get,
k + 3 = 3h + 6 –8
By further more simplification,
k = 3h -5…………………(i)
Now as we know that the line joining (2,3), (h,k) is perpendicular then we can say their product of slope is -1.
We know if a line is represented in the form of y = mx + c then its slope is m.
So, if y = 3x – 4, then its slope is 3.
If the points are $(x_1,y_1),(x_2,y_2)$ then the slope is ${\displaystyle \frac{y_2-y_1}{x_2-x_1}}$ .
So, for the points are (2,3) and (h,k), the slope is ${\displaystyle \frac{k-3}{h-2}}.$
Then now we can say,
${\displaystyle \frac{k-3}{h-2}}\times 3=-1$
Now by doing cross multiplication we can write it as,
3(k – 3) = -1(h – 2)
On further simplification we get,
3k – 9 = -h + 2
Now finally rearranging and simplifying we get,
3k + h = 1…………………………(ii)
In the equation (i) we got a relation k = 3h -5, so we can substitute it in equation (ii) we get,
3(3h – 5) + h = 11
So we can simplify it as,
9h – 15 + h = 11
Or, 10h = 15+11
Hence value of h is ${\displaystyle \frac{26}{10}}.$
Now substituting value of ‘h’ in equation (i) we get value of ‘K’ so,
k = 3h – 5
$\begin{array}{rl}& k=3\times {\displaystyle \frac{26}{10}}-5\\ & k={\displaystyle \frac{78-50}{10}}\\ & k={\displaystyle \frac{28}{10}}\end{array}$
So, $(h,k)=({\displaystyle \frac{26}{10}},{\displaystyle \frac{28}{10}})$ is the point image.
Now foot of perpendicular will be
$({\displaystyle \frac{h+2}{2}},{\displaystyle \frac{k+3}{2}})$
So by putting values $h={\displaystyle \frac{26}{10}}$ and $k={\displaystyle \frac{28}{10}}$ then foot of perpendicular will be
$\begin{array}{rl}& =({\displaystyle \frac{{\displaystyle \frac{26}{10}}+2}{2}},{\displaystyle \frac{{\displaystyle \frac{28}{10}}+3}{2}})\\ & =({\displaystyle \frac{{\displaystyle \frac{26+20}{10}}}{2}},{\displaystyle \frac{{\displaystyle \frac{28+30}{10}}}{2}})\\ & =({\displaystyle \frac{46}{20}},{\displaystyle \frac{58}{20}})\\ & =({\displaystyle \frac{23}{10}},{\displaystyle \frac{29}{10}})\end{array}$
So, the point of foot of perpendicular is $({\displaystyle \frac{23}{10}},{\displaystyle \frac{29}{10}}).$
The slope of point $(2,3),({\displaystyle \frac{23}{10}},{\displaystyle \frac{29}{10}})$ is
$m={\displaystyle \frac{{\displaystyle \frac{29}{10}}-3}{{\displaystyle \frac{23}{10}}-2}}={\displaystyle \frac{{\displaystyle \frac{29-30}{10}}}{{\displaystyle \frac{23-20}{10}}}}={\displaystyle \frac{-1}{3}}$
If point is $(x_1,y_1)$ and slope is m then we will use formula, $(y-y_1)=m(x-m_1)$
Using formula for point (2, 3) and slope ${\displaystyle \frac{-1}{3}}$ we get,
$y-3={\displaystyle \frac{-1}{3}}(x-2)$
On cross multiplication we get,
3(y – 3) = -x +2
So on simplification we get,
3y – 9 = -x + 2
Hence, x + 3y = 11
So, the foot of perpendicular is$({\displaystyle \frac{23}{10}},{\displaystyle \frac{29}{10}}).$
The Image is $({\displaystyle \frac{26}{10}},{\displaystyle \frac{28}{10}})$ and equation is x + 3y = 11.

Note: Students get confused how to find an image, so they should know that if a point is given and its image is taken any variable point then their mid-point will lie on the line equation as the distance between line and point will be the same as that of line and image.
Another way to find is first find the foot of perpendicular from the given point. Then use formula to find the image of the point. Then find the line equation by using two points line formula.