Question

exercise #3: given the points m, n, p, and q located at m(-4,1), n(0,4), p(-3,-3), and q(0,6)
(a) plot and label all four points.
(b) draw mn and pq.
(c) what seems to be true about the two lines you drew in (b)?
exercise #4: at what point or points would a circle with a radius of length 5 units and a center at (2,4) intersect a line that contains the points a(-3,8) and b(4,1). use the grid below, along with a ruler and compass, to help locate the point or points.
exercise #5: would the portion of ab contained inside of the circle be longer or shorter than the diameter of the circle? explain.
we will work with many quadrilaterals in the coordinate plane. many of these quadrilaterals will have parallel sides and perpendicular sides.
exercise #6: quadrilateral abcd has vertices located at a(4,8), b(8,2), c(-4,-6), and d(-8,0).
(a) plot the points and draw quadrilateral abcd.
(b) draw in ac and bd, the diagonals of the quadrilateral.
(c) what type of special quadrilateral does abcd appear to be?

Explanation:

Step1: Find the equation of the line passing through points $A(-3,8)$ and $B(4,1)$

The slope $m$ of the line passing through two - points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $m=\frac{y_2 - y_1}{x_2 - x_1}$. So, $m=\frac{1 - 8}{4+3}=\frac{-7}{7}=-1$.
Using the point - slope form $y - y_1=m(x - x_1)$ with the point $A(-3,8)$, we get $y - 8=-1(x + 3)$, which simplifies to $y=-x + 5$.

Step2: Find the equation of the circle

The standard form of a circle with center $(h,k)$ and radius $r$ is $(x - h)^2+(y - k)^2=r^2$. Here, $h = 2$, $k = 4$, and $r = 5$, so the equation of the circle is $(x - 2)^2+(y - 4)^2=25$.

Step3: Substitute $y=-x + 5$ into the equation of the circle

Substitute $y$ in the circle's equation: $(x - 2)^2+((-x + 5)-4)^2=25$.
Expand: $(x - 2)^2+(-x + 1)^2=25$.
$(x^{2}-4x + 4)+(x^{2}-2x + 1)=25$.
Combine like terms: $2x^{2}-6x+5 = 25$.
Rearrange to get a quadratic equation: $2x^{2}-6x - 20 = 0$.
Divide by 2: $x^{2}-3x - 10 = 0$.
Factor: $(x - 5)(x+2)=0$.
Solve for $x$: $x = 5$ or $x=-2$.

Step4: Find the corresponding $y$ - values

When $x = 5$, $y=-5 + 5=0$.
When $x=-2$, $y=-(-2)+5=7$.

Answer:

The intersection points are $(5,0)$ and $(-2,7)$.