The Straight Lines — MATHS STD 11 Science — Question
Gujarat BoardEnglish MediumSTD 11 ScienceMATHSThe Straight Lines4 Marks
Question
Find the equations to the altitudes of the triangle whose angular points are A(2, -2), B(1, 1) and C(-1, 0).
✓
Answer
AD,BE and CF are the three altitudes of the triangle We know, Slope of AD × Slope of BC = -1; AD passes through A(2, -2) Slope of BE × Slope of AC = -1; AD passes through A(1, 1) Slope of CF × Slope of AB = -1; AD passes through C(-1, 0) Slope of $\text{BC}=\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2} \Rightarrow$ Slope of AD = -2 Slope of $\text{AC}=\frac{0-(-2)}{-1-2}=\frac{2}{-3}=\frac{-2}{3}\Rightarrow$ Slope of $\text{BE}=\frac{3}{2}$ Slope of $\text{AB}=\frac{1+2}{1-2}=\frac{3}{-1}=-3\Rightarrow$ Slope of $\text{CF}=\frac{1}{3}$ So, for AD, we have $\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x})_1$$\Rightarrow\ \text{y}-(-2)=-2(\text{x}-2)$
$\Rightarrow\ \text{y}+2=-2\text{x}+4$
$\Rightarrow\ \text{2x}+\text{y}-2=0$
And, for BE, WE have
$\Rightarrow\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\Rightarrow\text{y}-1=\frac{3}{2}(\text{x}-1)$
$\Rightarrow\ \text{2y}-\text{3x}+1=0$
And, for CF, We have
$\text{y}-\text{y}=\text{m}(\text{x}-\text{x}_1)$$\Rightarrow\ \text{y}-0=\frac{1}{3}(\text{x}+1)$
$\Rightarrow\ \text{x}-\text{3y}+1=0$
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