If (2, 1), (–1, –2), (3, 3) are the midpoints of the sides BC, CA, AB of a triangle ABC, then equation of the line BC is
5x - 4y - 6 = 0
Understanding Triangle Side Midpoints
The problem provides us with the coordinates of the midpoints of the sides of a triangle ABC. Let these midpoints be:
- D = (2, 1) is the midpoint of side BC.
- E = (-1, -2) is the midpoint of side CA.
- F = (3, 3) is the midpoint of side AB.
Our goal is to determine the equation of the line that forms the side BC of this triangle.
Applying the Midpoint Theorem
A key concept in geometry is the Midpoint Theorem. It states that the line segment connecting the midpoints of any two sides of a triangle is parallel to the third side and is half the length of the third side.
In our triangle ABC:
- The line segment EF connects the midpoints of sides CA and AB. Therefore, EF is parallel to the third side, BC.
- Similarly, DE is parallel to AB, and FD is parallel to AC.
Since EF is parallel to BC, their slopes must be equal. We can use this property to find the slope of the line BC.
Calculating the Slope of Line BC
To find the slope of BC, we first calculate the slope of the line segment EF using the coordinates of E (-1, -2) and F (3, 3).
The formula for the slope ($m$) between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$Using the coordinates of E and F:
$$m_{EF} = \frac{3 - (-2)}{3 - (-1)}$$ $$m_{EF} = \frac{3 + 2}{3 + 1}$$ $$m_{EF} = \frac{5}{4}$$As established by the Midpoint Theorem, the line segment EF is parallel to the side BC. Therefore, the slope of BC ($m_{BC}$) is equal to the slope of EF ($m_{EF}$).
$$m_{BC} = m_{EF} = \frac{5}{4}$$Finding the Equation of Line BC
We now know the slope of the line BC is $\frac{5}{4}$. We also know that point D (2, 1) is the midpoint of BC, which means point D lies on the line BC.
We can use the point-slope form of the equation of a line, which is:
$$y - y_1 = m(x - x_1)$$Here, $(x_1, y_1) = (2, 1)$ and $m = \frac{5}{4}$. Substituting these values:
$$y - 1 = \frac{5}{4}(x - 2)$$To get the equation in the standard form ($Ax + By + C = 0$), we can rearrange the equation:
Multiply both sides by 4:
$$4(y - 1) = 5(x - 2)$$Distribute the numbers:
$$4y - 4 = 5x - 10$$Rearrange the terms to one side:
$$0 = 5x - 4y - 10 + 4$$ $$0 = 5x - 4y - 6$$Therefore, the equation of the line BC is $5x - 4y - 6 = 0$. This matches one of the given options.
Important Questions from General Equation of a Line
Two straight lines passing through the point A(3, 2) cut the line 2y = x + 3 and x-axis perpendicularly at P and Q respectively. The equation of the line PQ is
Lines x = ay + b, z = cy + d
and x = a'y + b', z = c'y + d'
are perpendicular, if
Equation of the line perpendicular to x - 2y = 1 and passing through (1, 1) is:
Find the equation of a straight line passing through (3, 4) and having slope 3.
If P(3, 4) is the mid-point of a line segment between the axes, then what is the equation of the line?
