Question

1.4 homework
question 23, 1.4.93
part 1 of 10
hw score: 84.62%, 22 of 26 points
points: 0 of 1
either prove or disprove the statement, \the points (-6, -4), (-2, -5), and (5,1) are the vertices of a right triangle.\ use a graph only as a guide.

let the three points (-6, -4), (-2, -5), and (5,1) be labeled as a, b, and c respectively. graph the triangle formed by the vertices a, b, and c. choose the correct graph below.

a. graph

b. graph

c. graph

d. graph

Explanation:

Step1: Recall the distance formula

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). We will find the lengths of \(AB\), \(BC\), and \(AC\).

Step2: Calculate \(AB\)

For \(A(-6,-4)\) and \(B(-2,-5)\):
\(x_1=-6,y_1 = - 4,x_2=-2,y_2=-5\)
\(AB=\sqrt{(-2 - (-6))^2+(-5 - (-4))^2}=\sqrt{(4)^2+(-1)^2}=\sqrt{16 + 1}=\sqrt{17}\)

Step3: Calculate \(BC\)

For \(B(-2,-5)\) and \(C(5,1)\):
\(x_1=-2,y_1=-5,x_2 = 5,y_2=1\)
\(BC=\sqrt{(5 - (-2))^2+(1 - (-5))^2}=\sqrt{(7)^2+(6)^2}=\sqrt{49+36}=\sqrt{85}\)

Step4: Calculate \(AC\)

For \(A(-6,-4)\) and \(C(5,1)\):
\(x_1=-6,y_1=-4,x_2 = 5,y_2=1\)
\(AC=\sqrt{(5 - (-6))^2+(1 - (-4))^2}=\sqrt{(11)^2+(5)^2}=\sqrt{121 + 25}=\sqrt{146}\)

Step5: Check Pythagorean theorem

We check if \(AB^{2}+BC^{2}=AC^{2}\), \(AB^{2}+AC^{2}=BC^{2}\) or \(BC^{2}+AC^{2}=AB^{2}\)
\(AB^{2}=17\), \(BC^{2}=85\), \(AC^{2}=146\)
\(17 + 85=102 eq146\)
\(17+146 = 163 eq85\)
\(85 + 146=231 eq17\)
Wait, maybe I made a mistake. Let's recalculate \(AB\), \(BC\), \(AC\) again.

Wait, recalculating \(AB\):
\(A(-6,-4)\), \(B(-2,-5)\)
\(\Delta x=-2-(-6)=4\), \(\Delta y=-5 - (-4)=-1\)
\(AB=\sqrt{4^{2}+(-1)^{2}}=\sqrt{16 + 1}=\sqrt{17}\) (correct)

\(BC\): \(B(-2,-5)\), \(C(5,1)\)
\(\Delta x=5-(-2)=7\), \(\Delta y=1-(-5)=6\)
\(BC=\sqrt{7^{2}+6^{2}}=\sqrt{49 + 36}=\sqrt{85}\) (correct)

\(AC\): \(A(-6,-4)\), \(C(5,1)\)
\(\Delta x=5-(-6)=11\), \(\Delta y=1-(-4)=5\)
\(AC=\sqrt{11^{2}+5^{2}}=\sqrt{121+25}=\sqrt{146}\) (correct)

Wait, maybe I labeled the points wrong. Wait, maybe the right angle is between other sides. Wait, let's use the slope formula. The slope of a line between \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\)

Slope of \(AB\): \(m_{AB}=\frac{-5-(-4)}{-2-(-6)}=\frac{-1}{4}=-\frac{1}{4}\)

Slope of \(BC\): \(m_{BC}=\frac{1-(-5)}{5-(-2)}=\frac{6}{7}\)

Slope of \(AC\): \(m_{AC}=\frac{1-(-4)}{5-(-6)}=\frac{5}{11}\)

Now, check if two slopes are negative reciprocals (product is -1)

\(m_{AB}\times m_{BC}=(-\frac{1}{4})\times\frac{6}{7}=-\frac{6}{28}=-\frac{3}{14} eq - 1\)

\(m_{AB}\times m_{AC}=(-\frac{1}{4})\times\frac{5}{11}=-\frac{5}{44} eq - 1\)

\(m_{BC}\times m_{AC}=\frac{6}{7}\times\frac{5}{11}=\frac{30}{77} eq - 1\)

Wait, but maybe I made a mistake in the problem. Wait, the points are \((-6,-4)\), \((-2,-5)\), \((5,1)\). Wait, let's recalculate the distances again.

Wait, \(AB\): distance between \((-6,-4)\) and \((-2,-5)\):

\((-2 - (-6)) = 4\), \((-5 - (-4))=-1\), so \(AB=\sqrt{4^2+(-1)^2}=\sqrt{17}\approx4.123\)

\(BC\): distance between \((-2,-5)\) and \((5,1)\):

\(5 - (-2)=7\), \(1 - (-5)=6\), so \(BC=\sqrt{7^2+6^2}=\sqrt{49 + 36}=\sqrt{85}\approx9.2195\)

\(AC\): distance between \((-6,-4)\) and \((5,1)\):

\(5 - (-6)=11\), \(1 - (-4)=5\), so \(AC=\sqrt{11^2+5^2}=\sqrt{121 + 25}=\sqrt{146}\approx12.083\)

Now, check \(AB^{2}+BC^{2}=17 + 85 = 102\), \(AC^{2}=146\), not equal. \(AB^{2}+AC^{2}=17+146 = 163\), \(BC^{2}=85\), not equal. \(BC^{2}+AC^{2}=85 + 146=231\), \(AB^{2}=17\), not equal. So the triangle is not a right triangle? But wait, maybe I made a mistake in the slope or distance.

Wait, wait, let's check the coordinates again. The points are \(A(-6,-4)\), \(B(-2,-5)\), \(C(5,1)\). Let's plot them mentally:

\(A\) is at x=-6, y=-4; \(B\) is at x=-2, y=-5 (so to the right of A, slightly down); \(C\) is at x=5, y=1 (to the right of B, up).

Wait, maybe I miscalculated the slope of \(AB\) and \(AC\) or \(BC\). Wait, let's recalculate the slope of \(AB\): \(\frac{-5 - (-4)}{-2 - (-6)}=\frac{-1}{4}\), correct. Slope of \(AC\): \(\frac{1 - (-4)}{5 - (-6)}=\frac{5}{11}\), correct. Slope of \(BC\): \(\frac{1 - (-5)}{5 - (-2)}=\frac{6}{7}\), correct.

Since the product of no two slopes is -1, the triangle is not a right triangle. But wait, the question first asks to choose the correct graph. Let's analyze the coordinates:

\(A(-6,-4)\): x=-6, y=-4 (third quadrant, close to x=-6, y=-4)

\(B(-2,-5)\): x=-2, y=-5 (fourth quadrant? No, x=-2 is left of y-axis, y=-5 is below x-axis, so third quadrant, x=-2, y=-5 (so to the right of A, down from A)

\(C(5,1)\): x=5, y=1 (first quadrant, right of y-axis, above x-axis)

So the graph should have A at (-6,-4), B at (-2,-5), C at (5,1). Looking at the options, option A has A and B in the lower left (third quadrant) and C in the first quadrant, which matches. So the correct graph is A.

Answer:

A