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2 years ago

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A parabola with the vertex at (1.5, 12.5) inscribes a segment of length 5 on x-axis. What is the area of the triangle formed by

points of intersection of parabola with the coordinate axes.

1 answer:

damaskus [11]2 years ago

5 0

Answer:

20 units²

Step-by-step explanation:

The x-intercepts are symmetrically located around the x-coordinate of the vertex, so are at

1.5 ± 5/2 = {-1, 4}

Using one of these we can find the unknown parameter "a" in the parabola's equation (in vertex form) ...

0 = a(4 -1.5)² +12.5

0 = 6.25a +12.5 . . . . . simplify

0 = a +2 . . . . . . . . . . . divide by 6.25

-2 = a

Then the standard-form equation of the parabola is ...

y = -2(x -1.5)² +12.5 = -2(x² -3x +2.25) +12.5

y = -2x² +6x +8

This tells us the y-intercept is 8. Then the relevant triangle has a base of 5 units and a height of 8. Its area is given by the formula ...

A = (1/2)bh = (1/2)(5)(8) = 20 . . . . units²

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. Given ????(5, −4) and T(−8,12):

damaskus [11]

Answer:

a) $y=\dfrac{13x}{16}-\dfrac{129}{16}$

b) $y = \dfrac{13x}{16}+ \dfrac{37}{2}$

Step-by-step explanation:

Given two points: $S(5,-4)$ and $T(-8,12)$

Since in both questions,a and b, we're asked to find lines that are perpendicular to ST. So, we'll do that first!

Perpendicular to ST:

the equation of any line is given by: $y = mx + c$ where, m is the slope(also known as gradient), and c is the y-intercept.

to find the perpendicular of ST <u>we first need to find the gradient of ST, using the gradient formula.</u>

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

the coordinates of S and T can be used here. (it doesn't matter if you choose them in any order: S can be either x_1 and y_1 or x_2 and y_2)

$m = \dfrac{12 - (-4)}{(-8) - 5}$

$m = \dfrac{-16}{13}$

to find the perpendicular of this gradient: we'll use:

$m_1m_2=-1$

both $m_1$ and $m_2$ denote slopes that are perpendicular to each other. So if $m_1 = \dfrac{12 - (-4)}{(-8) - 5}$ , then we can solve for $m_2$ for the slop of ther perpendicular!

$\left(\dfrac{-16}{13}\right)m_2=-1$

$m_2=\dfrac{13}{16}$ :: this is the slope of the perpendicular

a) Line through S and Perpendicular to ST

to find any equation of the line all we need is the slope $m$ and the points $(x,y)$ . And plug into the equation: $(y - y_1) = m(x-x_1)$

side note: you can also use the $y = mx + c$ to find the equation of the line. both of these equations are the same. but I prefer (and also recommend) to use the former equation since the value of 'c' comes out on its own.

$(y - y_1) = m(x-x_1)$

we have the slope of the perpendicular to ST i.e $m=\dfrac{13}{16}$

and the line should pass throught S as well, i.e $(5,-4)$ . Plugging all these values in the equation we'll get.

$(y - (-4)) = \dfrac{13}{16}(x-5)$

$y +4 = \dfrac{13x}{16}-\dfrac{65}{16}$

$y = \dfrac{13x}{16}-\dfrac{65}{16}-4$

$y=\dfrac{13x}{16}-\dfrac{129}{16}$

this is the equation of the line that is perpendicular to ST and passes through S

a) Line through T and Perpendicular to ST

we'll do the same thing for $T(-8,12)$

$(y - y_1) = m(x-x_1)$

$(y -12) = \dfrac{13}{16}(x+8)$

$y = \dfrac{13x}{16}+ \dfrac{104}{16}+12$

$y = \dfrac{13x}{16}+ \dfrac{37}{2}$

this is the equation of the line that is perpendicular to ST and passes through T

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Answer:

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Answer:

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Step-by-step explanation:

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