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Chapter 10: Problem 16

Find the coordinates of the foot of the perpendicular from the point \((2,-6)\) to the line \(3 y-x+2=0\).

Short Answer

Expert verified
(1/5, -3/5)

Step by step solution

01

- Understand the problem

Identify the given point \( (2, -6) \) and the line equation \( 3y - x + 2 = 0 \). The task is to find the coordinates of the foot of the perpendicular from the given point to the line.
02

- Find the slope of the given line

Rewrite the line equation in slope-intercept form \( y = mx + c \). Starting from \( 3y - x + 2 = 0 \), solve for \( y \): \[ 3y = x - 2 \] \[ y = \frac{1}{3} x - \frac{2}{3} \] This gives the slope of the line as \( m = \frac{1}{3} \).
03

- Determine the slope of the perpendicular line

The slope of the perpendicular line is the negative reciprocal of the original slope. Therefore, the slope of the perpendicular line is \(-3 \).
04

- Write the equation of the perpendicular line

Using the point-slope form \((y - y_1 = m(x - x_1))\) and the slope from Step 3, the equation of the line passing through \( (2, -6) \) with a slope of \(-3\) is: \[ y + 6 = -3(x - 2) \] Simplify this to get: \[ y = -3x + 6 - 6 \] \[ y = -3x + 0 \] \[ y = -3x \]
05

- Solve for the intersection point

Set the y-values of the two equations equal to each other to find the intersection point (foot of the perpendicular). \[ \frac{1}{3}x - \frac{2}{3} = -3x \] Multiply through by 3 to clear the fraction: \[ x - 2 = -9x \] Solve for \( x \): \[ x + 9x = 2 \] \[ 10x = 2 \] \[ x = \frac{2}{10} \] \[ x = \frac{1}{5} \]
06

- Find the corresponding y-coordinate

Substitute \( x = \frac{1}{5} \) back into the original line equation to find the corresponding \( y \) coordinate. \[ y = \frac{1}{3} \left( \frac{1}{5} \right) - \frac{2}{3} \] \[ y = \frac{1}{15} - \frac{2}{3} = \frac{1}{15} - \frac{10}{15} \] \[ y = \frac{1 - 10}{15} = \frac{-9}{15} \] \[ y = -\frac{3}{5} \]
07

- State the coordinates of the foot of the perpendicular

The coordinates of the foot of the perpendicular from the point \( (2, -6) \) to the line \( 3y - x + 2 = 0 \) are \( \left( \frac{1}{5}, -\frac{3}{5} \right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

line equation
To find the foot of the perpendicular from a point to a line, it is crucial to understand the line's equation. A line equation represents all the points along the line and can be written in various forms. In this task, we start with the equation in standard form: \[ 3y - x + 2 = 0 \]
To make it easier to work with, especially for finding slopes, we convert the standard form to the slope-intercept form \( y = mx + c \). This form shows the slope \( m \) and the y-intercept \( c \). Here's how we transform the given line equation:
Starting with \( 3y - x + 2 = 0 \), we solve for \( y \):
\[ 3y = x - 2 \]\[ y = \frac{1}{3} x - \frac{2}{3} \]
Now we see that the slope \( m \) is \( \frac{1}{3} \) and the y-intercept \( c \) is \( -\frac{2}{3} \).
slope
Understanding slope is essential to solving geometry problems involving lines. The slope \( m \) measures the steepness and direction of a line. In the slope-intercept form \( y = mx + c \), \( m \) is the coefficient of \( x \). A positive slope means the line rises as it moves from left to right, while a negative slope means it falls.
To find the slope of a perpendicular line, we use the negative reciprocal of the original slope. For the given line, with slope \( \frac{1}{3} \), the perpendicular slope is \( -3 \).
We use this new slope along with the coordinates of the given point \( (2, -6) \) to form the equation of the perpendicular line using the point-slope form: \( y - y_1 = m(x - x_1) \).
Thus, the perpendicular line equation is:
\[ y + 6 = -3(x - 2) \]
Simplifying, we get \( y = -3x \).
coordinates
Coordinates indicate the position of points in a plane. They are typically in the format \( (x, y) \). To find where two lines intersect, we set their equations equal to each other and solve for \( x \) and \( y \).
Here, we had:\[ \frac{1}{3}x - \frac{2}{3} = -3x \]
Multiplying through by 3 to remove the fractions, we get:\[ x - 2 = -9x \]Solving for \( x \) gives:\[ 10x = 2 \]\[ x = \frac{1}{5} \]Substituting \( x = \frac{1}{5} \) back into the original line equation \( y = \frac{1}{3}x - \frac{2}{3} \), we find \( y \):\[ y = \frac{1}{15} - \frac{10}{15} \]\[ y = -\frac{3}{5} \]Thus, the coordinates of the foot of the perpendicular from \( (2, -6) \) to the line \( 3y - x + 2 = 0 \) are \( \left( \frac{1}{5}, -\frac{3}{5} \right) \).

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Most popular questions from this chapter

The parabolas \(x^{2}=4 a y\) and \(y^{2}=4 a x\) meet at the origin and at the point \(P\). The tangent to \(x^{2}=4 a y\) at P meets \(y^{2}=4 a x\) again at \(A\), and the tangent to \(y^{2}=4 a x\) at P meets \(x^{2}=4 a y\) again at B. Prove that the angle \(\mathrm{APB}\) is \(\arctan \left(\frac{3}{4}\right)\) and that \(\mathrm{AB}\) is a common tangent to the two parabolas.

The circle \(\mathrm{S}_{1}\) with centre \(\mathrm{C}_{1}\left(a_{1}, b_{1}\right)\) and radius \(r_{1}\) touches externally the circle \(\mathrm{S}_{2}\) with centre \(\mathrm{C}_{2}\left(a_{2}, b_{2}\right)\) and radius \(r_{2}\). The tangent at their common point passes through the origin. Show that $$ \left(a_{1}{ }^{2}-a_{2}{ }^{2}\right)+\left(b_{1}{ }^{2}-b_{2}{ }^{2}\right)=\left(r_{1}{ }^{2}-r_{2}{ }^{2}\right) $$ If, also, the other two tangents from the origin to \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) are perpendicular, prove that \(\left|a_{2} b_{1}-a_{1} b_{2}\right|=\left|a_{1} a_{2}+b_{1} b_{2}\right|\) Hence show that, if \(C_{1}\) remains fixed but \(S_{1}\) and \(S_{2}\) vary, then \(C_{2}\) lies on the curve $$ \left(a_{1}{ }^{2}-b_{1}{ }^{2}\right)\left(x^{2}-y^{2}\right)+4 a_{1} b_{1} x y=0 $$

On the same diagram sketch the circles \(x^{2}+y^{2}=4\) and \(x^{2}+y^{2}-10 x=0\). The line \(a x+b y+1=0 \quad\) is a tangent to both these circles. State the distances of the centres of the circles from this tangent. Hence, or otherwise, find the possible values of \(a\) and \(b\) and show that, if \(2 \phi\). is the angle between the common tangents, then \(\tan \phi=\frac{3}{4}\).

Find the points of intersection of the circle $$ x^{2}+y^{2}-6 x+2 y-17=0 $$ nd the line \(x-y+2=0\). Show that an equation of the circle which has hese points as the ends of a diameter is $$ x^{2}+y^{2}-4 y-5=0 $$ Show also that this circle and the circle $$ x^{2}+y^{2}-8 x+2 y+13=0 $$ touch externally.

The parametric equations of a curve are \(x=\cos 2 t, \quad y=4 \sin t\). Sketch the curve for \(0 \leqslant t \leqslant \frac{1}{2} \pi\) Show that \(\frac{\mathrm{d} y}{\mathrm{~d} x}=-\operatorname{cosec} t \quad\) and find the equation of the tangent to the curve at the point \(\mathrm{A}(\cos 2 T, 4 \sin T\) ). The tangent at A crosses the \(x\)-axis at the point \(\mathrm{M}\) and the normal at \(\mathrm{A}\). crosses the \(x\)-axis at the point \(\mathrm{N}\). If the area of the triangle \(\mathrm{AMN}\) is \(12 \sin T\), find the value of \(T\) between 0 and \(\frac{1}{2} \pi\). \((\mathrm{AEB})^{\prime} 76\)
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