IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Geometry-Gradient of parallel lines
Topic :Geometry- Weightage : 21 %
All Questions for Topic : Metric conversions,Volume of regular polyhedra,Similarity and congruence,Coordinate geometry, including distance, midpoint and gradient formulae,Movement on a plane- isometric transformations, enlargements and tessellations,y=m x+c, gradients and intercepts (see also functions),Gradient of parallel lines,Circle geometry,Rotation around a given point
Question(6 Marks)
A line which passes through the point $(2,6)$ is perpendicular to $y=-4 x+5$.
a. Find the gradient of the given line.
b. Find the gradient of this perpendicular line.
c. State the equation of the perpendicular line.
▶️Answer/Explanation
Ans:
a. Find the gradient of the given line:
The given line is \(y = -4x + 5\), and this equation is in the slope-intercept form, where the coefficient of \(x\) (-4 in this case) represents the gradient.
So, the gradient of the given line is \(m_1 = -4\).
b. Find the gradient of the perpendicular line:
Let’s call the gradient of the perpendicular line \(m_2\) (to be determined). Since the given line and the perpendicular line are perpendicular, their gradients are related by the following equation:
\(m_1 \cdot m_2 = -1\)
Now, solve for \(m_2\):
\(m_2 = -\frac{1}{m_1} = -\frac{1}{-4} = \frac{1}{4}\)
So, the gradient of the perpendicular line is \(m_2 = \frac{1}{4}\).
c. State the equation of the perpendicular line:
Now that we know the gradient of the perpendicular line \(\left(\frac{1}{4}\right)\) and the point it passes through \((2, 6)\), we can use the point-slope form of a linear equation to write the equation of the perpendicular line:
\(y – y_1 = m_2(x – x_1)\)
where \((x_1, y_1)\) is the given point \((2, 6)\) and \(m_2\) is the gradient we found (\(\frac{1}{4}\)).
Now, plug in the values:
\(y – 6 = \frac{1}{4}(x – 2)\)
To put the equation in slope-intercept form (\(y = mx + b\)), let’s simplify:
\(y – 6 = \frac{1}{4}x – \frac{1}{2}\)
Now, isolate \(y\):
\(y = \frac{1}{4}x + \frac{11}{2}\)
So, the equation of the perpendicular line passing through the point \((2, 6)\) is \(y = \frac{1}{4}x + \frac{11}{2}\).