GMAT Quantitative - Coordinate Geometry

Determine the equation of the tangent line to the following curve at the point :

Explanation

First find the slope of the tangent line by taking the derivative of the function and plugging in the x value of the given point to find the slope of the curve at that location:

So the slope of the tangent line to the curve at the given point is . The next step is to plug this slope into the formula for a line, along with the coordinates of the given point, to solve for the value of the y intercept of the tangent line:

$-7=1(2)+b\rightarrow b=-9$

We now know the slope and y intercept of the tangent line, so we can write its equation as follows:

Give the slope of the line of the equation:

$\frac{4}{5}$

$-\frac{4}{5}$

$\frac{5}{4}$

$-\frac{5}{4}$

Explanation

Rewrite in the slope-intercept form :

$\left (0.4x - 0.5y \right )\cdot 10 = 20 \cdot 10$

$\left (0.4x \right )\cdot 10-\left ( 0.5y \right )\cdot 10 = 200$

$-5y \div (-5)=\left ( -4x+ 200 \right )\div (-5)$

$y=\left ( -4x \right ) \div (-5)+\left ( 200 \right ) \div (-5)$

$y= \frac{4}{5}x-40$

The slope is the coefficient of , which is $\frac{4}{5}$ .

What is distance between and ?

$\sqrt{29}$

$\sqrt{12}$

$\sqrt{7}$

$\sqrt{20}$

Explanation

$distance= \sqrt{(x_{2}-x_{1})^2+y_{2}-y_{1})^2}$

$= \sqrt{(6-1)^2+(7-5)^2}$

$= \sqrt{(5)^2+(2)^2}$

$= \sqrt{(25+4)}$

$\sqrt{29}$

Suppose the points and are plotted to connect a line. What are the -intercept and -intercept, respectively?

$-\frac{7}{5},\frac{7}{4}$

$\frac{7}{5},\frac{7}{4}$

$-\frac{5}{4},\frac{7}{4}$

$-\frac{4}{5},\frac{7}{4}$

Explanation

First, given the two points, find the equation of the line using the slope formula and the y-intercept equation.

Slope:

$\frac{y_2-y_1}{x_2-x_1} = \frac{3-(-2)}{1-(-3)}=\frac{5}{4}$

Write the slope-intercept formula.

Substitute a given point and the slope into the equation to find the y-intercept.

$3=\left(\frac{5}{4}\right)(1)+b$

$\frac{12}{4}=\frac{5}{4}+b$

$b=\frac{7}{4}$

The y-intercept is: $b=\frac{7}{4}$ .

Substitiute the slope and the y-intercept into the slope-intercept form.

$y=\frac{5}{4}x+\frac{7}{4}$

To find the x-intercept, substitute and solve for x.

$0=\frac{5}{4}x+\frac{7}{4}$

$-\frac{7}{4}=\frac{5}{4}x$

$-\frac{7}{4} \cdot (\frac{4}{5})=\frac{5}{4}x \cdot (\frac{4}{5})$

$x=-\frac{7}{5}$

The x-intercept is: $x=-\frac{7}{5}$

In the -plane, point $(a,b)$ lies on a circle with center at the origin. The radius of the circle is 5. What is the value of $a^{2}+b^{2}$ ?

$25$

$5$

$10$

$16$

$32$

Explanation

$a$ and $b$ are the right-angle sides of a triangle, and the radius of the circle is the hypotenuse of the triangle. From the Pythagorean Theorem we would know that $a^{2}+b^{2}=r^{2}=5^{2}=25$ .

In the -plane, point $(a,b)$ lies on a circle with center at the origin. The radius of the circle is 5. What is the value of $a^{2}+b^{2}$ ?

$25$

$5$

$10$

$16$

$32$

Explanation

$a$ and $b$ are the right-angle sides of a triangle, and the radius of the circle is the hypotenuse of the triangle. From the Pythagorean Theorem we would know that $a^{2}+b^{2}=r^{2}=5^{2}=25$ .

A line segment has its midpoint at and an endpoint at . What are the coordinates of the other endpoint?

Explanation

Because we are given the midpoint and one of the endpoints, we know the x coordinate of the other endpoint will be the same distance away from the midpoint in the x direction, and the y coordinate of the other endpoint will be the same distance away from the midpoint in the y direction. Given two endpoints of the form:

The midpoint of these two endpoints has the coordinates:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Plugging in values for the given midpoint and one of the endpoints, which we can see is because it lies to the right of the midpoint, we can solve for the other endpoint as follows:

$(x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(-1,3)=(\frac{x_1+3}{2},\frac{y_1+8}{2})$

$\frac{x_1+3}{2}=-1\rightarrow x_1+3=-2\rightarrow x_1=-5$

$\frac{y_1+8}{2}=3\rightarrow y_1+8=6\rightarrow y_1=-2$

So the other endpoint has the coordinates

Give the slope of the line of the equation:

$\frac{4}{5}$

$-\frac{4}{5}$

$\frac{5}{4}$

$-\frac{5}{4}$

Explanation

Rewrite in the slope-intercept form :

$\left (0.4x - 0.5y \right )\cdot 10 = 20 \cdot 10$

$\left (0.4x \right )\cdot 10-\left ( 0.5y \right )\cdot 10 = 200$

$-5y \div (-5)=\left ( -4x+ 200 \right )\div (-5)$

$y=\left ( -4x \right ) \div (-5)+\left ( 200 \right ) \div (-5)$

$y= \frac{4}{5}x-40$

The slope is the coefficient of , which is $\frac{4}{5}$ .

Determine the equation of the tangent line to the following curve at the point :

Explanation

First find the slope of the tangent line by taking the derivative of the function and plugging in the x value of the given point to find the slope of the curve at that location:

$-7=1(2)+b\rightarrow b=-9$

We now know the slope and y intercept of the tangent line, so we can write its equation as follows:

What is distance between and ?

$\sqrt{29}$

$\sqrt{12}$

$\sqrt{7}$

$\sqrt{20}$

Explanation

$distance= \sqrt{(x_{2}-x_{1})^2+y_{2}-y_{1})^2}$

$= \sqrt{(6-1)^2+(7-5)^2}$

$= \sqrt{(5)^2+(2)^2}$

$= \sqrt{(25+4)}$

$\sqrt{29}$

Coordinate Geometry

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