Find the equation of a straight line passing through (3, 4) and having slope 3.
This problem requires us to find the equation of a straight line given specific information: a point it passes through and its slope.
The most direct way to find the equation of a line when you have a point and the slope is by using the point-slope form. The formula is:
$$y - y_1 = m(x - x_1)$$
Where:
Let's substitute the given values into the point-slope formula:
Now, we compare our derived equation, $3x - y - 5 = 0$, with the given options:
Our calculated equation perfectly matches Option 1.
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:
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
If P(3, 4) is the mid-point of a line segment between the axes, then what is the equation of the line?