Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 178374 by depressiveshrek last updated on 16/Oct/22

Let the points ABC form a triangle on the cartesian plane, whose area is 20. Let the coordinates of said points be A(8, 6) B(2, 4) and C(x, y) If ∣AC∣=∣BC∣, find the coordinates of point C.

$${Let}\:{the}\:{points}\:{ABC}\:{form}\:{a}\:{triangle}\:{on}\:{the} \\ $$$${cartesian}\:{plane},\:{whose}\:{area}\:{is}\:\mathrm{20}.\:{Let}\:{the}\:{coordinates} \\ $$$${of}\:{said}\:{points}\:{be}\:{A}\left(\mathrm{8},\:\mathrm{6}\right)\:{B}\left(\mathrm{2},\:\mathrm{4}\right)\:{and}\:{C}\left({x},\:{y}\right) \\ $$$${If}\:\mid{AC}\mid=\mid{BC}\mid,\:{find}\:{the}\:{coordinates}\:{of}\:{point}\:{C}. \\ $$

Answered by Ar Brandon last updated on 16/Oct/22

Let M be the midpoint of BC. Then M has coordinates (((8+2)/2), ((6+4)/2))=M(5, 5) If ∣AC∣=∣BC∣ then Δ_(ABC) is an isosceles triangle ∣AM∣=(√((8−5)^2 +(6−5)^2 ))=(√(10)) ∣CM∣=(√((x−5)^2 +(y−5)^2 )) ∣AC∣=∣BC∣ ⇒(√((x−8)^2 +(y−6)^2 ))=(√((x−2)^2 +(y−4)^2 )) ⇒−6(2x−10)=2(2y−10) ⇒y−5=−3(x−5) ⇒∣CM∣=(√((x−5)^2 +9(x−5)^2 ))=(√(10))∣x−5∣ Area of triangle=20 ⇒(1/2)∣AM∣∙∣CM∣=20 ⇒(1/2)(√(10))×(√(10))∣x−5∣=10 ⇒x=7 ⇒y−5=−3(7−5) ⇒y=−1 Hence coordinates of C are C(7, −1)

$$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{BC}.\:\mathrm{Then} \\ $$$${M}\:\mathrm{has}\:\mathrm{coordinates}\:\left(\frac{\mathrm{8}+\mathrm{2}}{\mathrm{2}},\:\frac{\mathrm{6}+\mathrm{4}}{\mathrm{2}}\right)={M}\left(\mathrm{5},\:\mathrm{5}\right) \\ $$$$\mathrm{If}\:\mid{AC}\mid=\mid{BC}\mid\:\mathrm{then}\:\Delta_{{ABC}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle} \\ $$$$\mid{AM}\mid=\sqrt{\left(\mathrm{8}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{6}−\mathrm{5}\right)^{\mathrm{2}} }=\sqrt{\mathrm{10}} \\ $$$$\mid{CM}\mid=\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−\mathrm{5}\right)^{\mathrm{2}} } \\ $$$$\mid{AC}\mid=\mid{BC}\mid \\ $$$$\Rightarrow\sqrt{\left({x}−\mathrm{8}\right)^{\mathrm{2}} +\left({y}−\mathrm{6}\right)^{\mathrm{2}} }=\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow−\mathrm{6}\left(\mathrm{2}{x}−\mathrm{10}\right)=\mathrm{2}\left(\mathrm{2}{y}−\mathrm{10}\right)\:\Rightarrow{y}−\mathrm{5}=−\mathrm{3}\left({x}−\mathrm{5}\right) \\ $$$$\Rightarrow\mid{CM}\mid=\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{9}\left({x}−\mathrm{5}\right)^{\mathrm{2}} }=\sqrt{\mathrm{10}}\mid{x}−\mathrm{5}\mid \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{triangle}=\mathrm{20} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mid{AM}\mid\centerdot\mid{CM}\mid=\mathrm{20}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{10}}×\sqrt{\mathrm{10}}\mid{x}−\mathrm{5}\mid=\mathrm{10} \\ $$$$\Rightarrow{x}=\mathrm{7}\:\Rightarrow{y}−\mathrm{5}=−\mathrm{3}\left(\mathrm{7}−\mathrm{5}\right)\:\Rightarrow{y}=−\mathrm{1} \\ $$$$\mathrm{Hence}\:\mathrm{coordinates}\:\mathrm{of}\:{C}\:\mathrm{are}\:{C}\left(\mathrm{7},\:−\mathrm{1}\right) \\ $$

Commented by depressiveshrek last updated on 16/Oct/22

Correct, but there is also another point of C that helps to form the triangle.

$${Correct},\:{but}\:{there}\:{is}\:{also}\:{another}\:{point}\:{of}\:{C}\:{that} \\ $$$${helps}\:{to}\:{form}\:{the}\:{triangle}. \\ $$

Commented by Ar Brandon last updated on 16/Oct/22

Commented by Ar Brandon last updated on 16/Oct/22

The second point is the reflection of C_1 on y_(AB)

$$\mathrm{The}\:\mathrm{second}\:\mathrm{point}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reflection}\:\mathrm{of}\:{C}_{\mathrm{1}} \:\mathrm{on}\:\mathrm{y}_{{AB}} \\ $$

Commented by Ar Brandon last updated on 16/Oct/22

Midpoint of C_1 C_2 =M(5, 5) ((x+7)/2)=5, ((y−1)/2)=5 ⇒x=3, y=11 C_2 (x, y)=C_2 (3, 11)

$$\mathrm{Midpoint}\:\mathrm{of}\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} ={M}\left(\mathrm{5},\:\mathrm{5}\right) \\ $$$$\frac{{x}+\mathrm{7}}{\mathrm{2}}=\mathrm{5},\:\frac{\mathrm{y}−\mathrm{1}}{\mathrm{2}}=\mathrm{5} \\ $$$$\Rightarrow{x}=\mathrm{3},\:\mathrm{y}=\mathrm{11} \\ $$$${C}_{\mathrm{2}} \left({x},\:\mathrm{y}\right)={C}_{\mathrm{2}} \left(\mathrm{3},\:\mathrm{11}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com