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Question Number 8608 by tawakalitu last updated on 17/Oct/16

if any angle of equilateral triangle is (− 1, 2) and any one side is x − (√(3y)) + 5 = 0 then equation of the other two sides are ??

$$\mathrm{if}\:\mathrm{any}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{is} \\ $$$$\left(−\:\mathrm{1},\:\mathrm{2}\right)\:\mathrm{and}\:\mathrm{any}\:\mathrm{one}\:\mathrm{side}\:\mathrm{is}\:\:\mathrm{x}\:−\:\sqrt{\mathrm{3y}}\:+\:\mathrm{5}\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{are}\:?? \\ $$

Answered by sandy_suhendra last updated on 18/Oct/16

Answered by sandy_suhendra last updated on 18/Oct/16

let △ABC is the equilateral triangle the equation of BC^ ⇒ x−(√3) y+5=0 so the gradient of BC=m_(BC) =tan ∠BPQ=(1/3)(√3) ⇒∠BPQ=30° ∠ACB=60°⇒∠PCA=180°−60°=120° ∠CRQ=∠BPQ+∠PCA=30°+120°=150° (AC intercepts X axis at R) the gradient of AC=tan ∠CRQ=tan 150°=−(1/3)(√3) the equation of AC ⇒ y=m_(AC) (x−x_A )+y_A y=−(1/3)(√3)(x+1)+2 ∠BQP=180°−∠BPQ−∠CBA=180°−30°−60°=90°=∠BQX the gradient of AB=tan ∠BQX=tan 90°=∞ the equation of AB ⇒ x=−1

$$\mathrm{let}\:\bigtriangleup\mathrm{ABC}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{B}\overset{} {\mathrm{C}}\:\Rightarrow\:\mathrm{x}−\sqrt{\mathrm{3}}\:\mathrm{y}+\mathrm{5}=\mathrm{0}\: \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{gradient}\:\mathrm{of}\:\mathrm{BC}=\mathrm{m}_{\mathrm{BC}} =\mathrm{tan}\:\angle\mathrm{BPQ}=\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{3}}\:\Rightarrow\angle\mathrm{BPQ}=\mathrm{30}° \\ $$$$\angle\mathrm{ACB}=\mathrm{60}°\Rightarrow\angle\mathrm{PCA}=\mathrm{180}°−\mathrm{60}°=\mathrm{120}° \\ $$$$\angle\mathrm{CRQ}=\angle\mathrm{BPQ}+\angle\mathrm{PCA}=\mathrm{30}°+\mathrm{120}°=\mathrm{150}° \\ $$$$\left(\mathrm{AC}\:\mathrm{intercepts}\:\mathrm{X}\:\mathrm{axis}\:\mathrm{at}\:\mathrm{R}\right) \\ $$$$\mathrm{the}\:\mathrm{gradient}\:\mathrm{of}\:\mathrm{AC}=\mathrm{tan}\:\angle\mathrm{CRQ}=\mathrm{tan}\:\mathrm{150}°=−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{3}} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{AC}\:\Rightarrow\:\mathrm{y}=\mathrm{m}_{\mathrm{AC}} \left(\mathrm{x}−\mathrm{x}_{\mathrm{A}} \right)+\mathrm{y}_{\mathrm{A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{3}}\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{2} \\ $$$$\angle\mathrm{BQP}=\mathrm{180}°−\angle\mathrm{BPQ}−\angle\mathrm{CBA}=\mathrm{180}°−\mathrm{30}°−\mathrm{60}°=\mathrm{90}°=\angle\mathrm{BQX} \\ $$$$\mathrm{the}\:\mathrm{gradient}\:\mathrm{of}\:\mathrm{AB}=\mathrm{tan}\:\angle\mathrm{BQX}=\mathrm{tan}\:\mathrm{90}°=\infty \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{AB}\:\Rightarrow\:\mathrm{x}=−\mathrm{1} \\ $$

Commented by tawakalitu last updated on 18/Oct/16

Thank you for your help.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{help}. \\ $$

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