In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.
Solution and Explanation
Let AD be the altitude of triangle ABC from vertex A.
Accordingly, \(AD⊥BC\)
The equation of the line passing through point (2, 3) and having a slope of 1 is
\((y - 3) = 1(x - 2)\)
\(⇒ x- y + 1 = 0\)
\(⇒ y - x = 1\)
Therefore, equation of the altitude from vertex \(A = y - x = 1\).
Length of AD = Length of the perpendicular from A (2, 3) to BC
The equation of BC is
\((y+1)=\frac{2+1}{1-4}(x-4)\)
\(⇒(y+1)=-1(x-4)\)
\(⇒y+1=-x+4\)
\(⇒x+y-3=0.......(1)\)
The perpendicular distance (d) of a line \(Ax + By + C = 0\) from a point \((x_1, y_1)\) is given by
\(d=\frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^2+B^2}}\)
On comparing equation (1) to the general equation of line \(Ax + By + C = 0\), we obtain \(A = 1, B = 1\), and \(C = -3.\)
∴ Length of \(AD=\frac{\left|1\times2+1\times3-3\right|}{\sqrt{1^2+1^2}}\) units
\(=\frac{\left|2\right|}{\sqrt2}\) units
\(=\frac{2}{\sqrt2}\) units
\(=\sqrt2\) units
Thus, the equation and the length of the altitude from vertex A are \(y - x = 1\) and \(\sqrt2\) units respectively.
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c