Answered

Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Ask your questions and receive precise answers from experienced professionals across different disciplines. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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 :

brainsoft

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]

We hope this was helpful. Please come back whenever you need more information or answers to your queries. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.