Solved on Nov 20, 2023

Find the distance between the line $y = x - 4$ and the point $(-1, 5)$ . Round the answer to the nearest tenth.

STEP 1

Assumptions1. The equation of line $\ell$ is $y=x-4$ . The point $Y$ has coordinates $(-1,5)$ 3. We need to find the shortest distance between line $\ell$ and point $Y$ , which is the perpendicular distance from the point to the line

STEP 2

The formula to find the distance $d$ from a point $(x0, y0)$ to a line $ax + by + c =0$ is given by $d = \frac{|ax0 + by0 + c|}{\sqrt{a^2 + b^2}}$

STEP 3

We first need to rewrite the equation of line $\ell$ in the form $ax + by + c =0$ . The equation $y=x-$ can be rewritten as $x-y-=0$ .
So, $a=1$ , $b=-1$ , and $c=-$ .

STEP 4

Now, we can substitute the values of $a$ , $b$ , $c$ , and the coordinates of point $Y$ into the distance formula.
$d = \frac{|1(-1) -1() -4|}{\sqrt{1^2 + (-1)^2}}$

STEP 5

implify the numerator and the denominator separately.
$d = \frac{|-1 -5 -4|}{\sqrt{1 +1}}$

STEP 6

Further simplify the expression.
$d = \frac{|-10|}{\sqrt{2}}$

STEP 7

Calculate the absolute value in the numerator and simplify the denominator.
$d = \frac{10}{\sqrt{2}}$

STEP 8

Rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{2}$ .
$d = \frac{10\sqrt{2}}{2}$

STEP 9

implify the expression.
$d =5\sqrt{2}$

STEP 10

Now, we need to round our answer to the nearest tenth. $d \approx7.$ So, the distance between line $\ell$ and the point $Y(-,5)$ is approximately7. units.

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Find the distance between the line $y = x - 4$ and the point $(-1, 5)$ . Round the answer to the nearest tenth.

STEP 1

Assumptions1. The equation of line $\ell$ is $y=x-4$ . The point $Y$ has coordinates $(-1,5)$ 3. We need to find the shortest distance between line $\ell$ and point $Y$ , which is the perpendicular distance from the point to the line

STEP 2

The formula to find the distance $d$ from a point $(x0, y0)$ to a line $ax + by + c =0$ is given by $d = \frac{|ax0 + by0 + c|}{\sqrt{a^2 + b^2}}$

STEP 3

We first need to rewrite the equation of line $\ell$ in the form $ax + by + c =0$ . The equation $y=x-$ can be rewritten as $x-y-=0$ .
So, $a=1$ , $b=-1$ , and $c=-$ .

STEP 4

Now, we can substitute the values of $a$ , $b$ , $c$ , and the coordinates of point $Y$ into the distance formula.
$d = \frac{|1(-1) -1() -4|}{\sqrt{1^2 + (-1)^2}}$

STEP 5

implify the numerator and the denominator separately.
$d = \frac{|-1 -5 -4|}{\sqrt{1 +1}}$

STEP 6

Further simplify the expression.
$d = \frac{|-10|}{\sqrt{2}}$

STEP 7

Calculate the absolute value in the numerator and simplify the denominator.
$d = \frac{10}{\sqrt{2}}$

STEP 8

Rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{2}$ .
$d = \frac{10\sqrt{2}}{2}$

STEP 9

implify the expression.
$d =5\sqrt{2}$

STEP 10

Now, we need to round our answer to the nearest tenth. $d \approx7.$ So, the distance between line $\ell$ and the point $Y(-,5)$ is approximately7. units.

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Find the distance between the line y=x−4y = x - 4y=x−4 and the point (−1,5)(-1, 5)(−1,5). Round the answer to the nearest tenth.

STEP 1

Assumptions1. The equation of line ℓ\ellℓ is y=x−4y=x-4y=x−4 . The point YYY has coordinates (−1,5)(-1,5)(−1,5)3. We need to find the shortest distance between line ℓ\ellℓ and point YYY, which is the perpendicular distance from the point to the line

STEP 2

The formula to find the distance ddd from a point (x0,y0)(x0, y0)(x0,y0) to a line ax+by+c=0ax + by + c =0ax+by+c=0 is given byd=∣ax0+by0+c∣a2+b2d = \frac{|ax0 + by0 + c|}{\sqrt{a^2 + b^2}}d=a2+b2​∣ax0+by0+c∣​

STEP 3

We first need to rewrite the equation of line ℓ\ellℓ in the form ax+by+c=0ax + by + c =0ax+by+c=0. The equation y=x−y=x-y=x− can be rewritten as x−y−=0x-y-=0x−y−=0.So, a=1a=1a=1, b=−1b=-1b=−1, and c=−c=-c=−.

STEP 4

Now, we can substitute the values of aaa, bbb, ccc, and the coordinates of point YYY into the distance formula.d=∣1(−1)−1()−4∣12+(−1)2d = \frac{|1(-1) -1() -4|}{\sqrt{1^2 + (-1)^2}}d=12+(−1)2​∣1(−1)−1()−4∣​

STEP 5

implify the numerator and the denominator separately.d=∣−1−5−4∣1+1d = \frac{|-1 -5 -4|}{\sqrt{1 +1}}d=1+1​∣−1−5−4∣​

STEP 6

Further simplify the expression.d=∣−10∣2d = \frac{|-10|}{\sqrt{2}}d=2​∣−10∣​

STEP 7

Calculate the absolute value in the numerator and simplify the denominator.d=102d = \frac{10}{\sqrt{2}}d=2​10​

STEP 8

Rationalize the denominator by multiplying the numerator and the denominator by 2\sqrt{2}2​.d=1022d = \frac{10\sqrt{2}}{2}d=2102​​

STEP 9

implify the expression.d=52d =5\sqrt{2}d=52​

STEP 10

Now, we need to round our answer to the nearest tenth.d≈7.d \approx7.d≈7.So, the distance between line ℓ\ellℓ and the point Y(−,5)Y(-,5)Y(−,5) is approximately7. units.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the distance between the line $y = x - 4$ and the point $(-1, 5)$ . Round the answer to the nearest tenth.

Assumptions1. The equation of line $\ell$ is $y=x-4$ . The point $Y$ has coordinates $(-1,5)$ 3. We need to find the shortest distance between line $\ell$ and point $Y$ , which is the perpendicular distance from the point to the line

The formula to find the distance $d$ from a point $(x0, y0)$ to a line $ax + by + c =0$ is given by $d = \frac{|ax0 + by0 + c|}{\sqrt{a^2 + b^2}}$

We first need to rewrite the equation of line $\ell$ in the form $ax + by + c =0$ . The equation $y=x-$ can be rewritten as $x-y-=0$ .
So, $a=1$ , $b=-1$ , and $c=-$ .

Now, we can substitute the values of $a$ , $b$ , $c$ , and the coordinates of point $Y$ into the distance formula.
$d = \frac{|1(-1) -1() -4|}{\sqrt{1^2 + (-1)^2}}$

implify the numerator and the denominator separately.
$d = \frac{|-1 -5 -4|}{\sqrt{1 +1}}$

Further simplify the expression.
$d = \frac{|-10|}{\sqrt{2}}$

Calculate the absolute value in the numerator and simplify the denominator.
$d = \frac{10}{\sqrt{2}}$

Rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{2}$ .
$d = \frac{10\sqrt{2}}{2}$

implify the expression.
$d =5\sqrt{2}$

Now, we need to round our answer to the nearest tenth. $d \approx7.$ So, the distance between line $\ell$ and the point $Y(-,5)$ is approximately7. units.