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Given points $A(2,3)$, $B(-1,4)$, and $C(-2,-2)$, determine point $D$ so that the slope from $A$ to $B$ equals the slope from $C$ to $D$, and the slope from $A$ to $D$ equals the slope from $B$ to $C$.

Sagot :

Brainsoftidn

Answer:

D is approximately (-2, -1)

Step-by-step explanation:

[tex]{ \tt{slope \:AB = slope \: CD }} \\ \\ { \tt{ \frac{(4 - 3)}{( - 1 - 2)} = \frac{(y - ( - 2))}{(x - ( - 2))} }} \\ \\ { \tt{ \frac{1}{ - 3} = \frac{y + 2}{x + 2} }} \\ \\ { \tt{x + 2 = - 3y - 6}} \\ { \underline{ \tt{ \green{ \: \: 3y + x = - 8 \: \: }}}}[/tex]

[tex]{ \tt{slope \: AD= slope \: BC}} \\ \\ { \tt{ \frac{y - 3}{x - 2} = \frac{ - 2 - 4}{ - 2 - 1} }} \\ \\ { \tt{ \frac{y - 3}{x - 2} = \frac{6}{3} }} \\ \\ { \tt{ \frac{y - 3}{x - 2} = 2}} \\ \\ { \tt{y - 3 = 2(x - 2)}} \\ { \tt{y - 3 = 2x - 4}} \\ { \underline{ \tt{ \blue{ \: \: y - 2x = - 1 \: \: }}}}[/tex]

Solve the green equation and blue equation simultaneously:

[tex]{ \boxed{ \tt{ \red{ \: y \approx - 2 \: \: }}and \: \: { \red{x \approx - 1}}}}[/tex]

Let the co-ordinates of D be (a,b)

  • Slope of AB =Slope of CD

[tex]\\ \tt\hookrightarrow \dfrac{4-3}{-1-2}=\dfrac{b+2}{a+2}[/tex]

[tex]\\ \tt\hookrightarrow \dfrac{-1}{2}=\dfrac{b+2}{a+2}[/tex]

[tex]\\ \tt\hookrightarrow -a-2=2b+4[/tex]

[tex]\\ \tt\hookrightarrow a+2b+6=0\dots(1)[/tex]

  • Slope of AD=Slope of BC

[tex]\\ \tt\hookrightarrow \dfrac{b-3}{a-2}=\dfrac{-2-4}{-2+1}[/tex]

[tex]\\ \tt\hookrightarrow \dfrac{b-3}{a-2}=6[/tex]

[tex]\\ \tt\hookrightarrow 6a-12=b-3[/tex]

[tex]\\ \tt\hookrightarrow 6a-b-9=0\dots(2)[/tex]

Multiplying 2 with eq(2)

[tex]\\ \tt\hookrightarrow 12a-2b-18=0\dots(3)[/tex]

  • Add eq(1) and (3)

[tex]\\ \tt\hookrightarrow 13a-12=0[/tex]

[tex]\\ \tt\hookrightarrow a=12/13=0.9\to 1[/tex]

  • Put in eq(1)

[tex]\\ \tt\hookrightarrow 12/13+2b+6=0[/tex]

[tex]\\ \tt\hookrightarrow 90/13=-2b[/tex]

[tex]\\ \tt\hookrightarrow b=-90/26=-3 4\to 3[/tex]

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