Question 1
Topic: Topic : SL 3.1 and Topic: SL 2.1
Given :
- Coordinate of Point A (-3, 2) and Coordinate of Point B (15, -8)
- N Mid point of line segment AB
- Line L is perpendicular to line segment [AB] and intersects the midpoint N
Find:
- Coordinate of N
- slope of L
- equation of line L in form of y=mx +c.
▶️Answer/Explanation
Answer:
(a)
The coordinates of the midpoint N can be found by taking the average of the x-coordinates and y-coordinates of points A and B.
The x-coordinate of \(N (N_{x})\)is the average of the x-coordinates of A and B:
\(N_{x}=\frac{\left ( -3+15 \right )}{2}=\frac{12}{2}=6\)
The y-coordinate of \(N (N_{y})\) is the average of the y-coordinates of A and B
\(N_{y}=\frac{\left ( 2+\left ( -8 \right ) \right )}{2}=-\frac{6}{2}=-3\)
So, the coordinates of point N are (6, -3).
(b)
The slope of the line segment [AB] can be found using the formula:
\(Slpoe(m)=\frac{Change in y}{Change in x}\)
\(m=\frac{\left ( -8-2 \right )}{15-\left ( -3 \right )}=\frac{-10}{18}=-\frac{5}{9}\)
The slope of a line perpendicular to this line (let’s call it L) is the negative reciprocal of the slope of AB. Therefore, the slope of line L is \(\frac{9}{5}\)
(c)
Now, we know the slope of line L\( \frac{9}{5}\)and the coordinates of the midpoint N (6, -3). We can use the point-slope form of the equation of a line:
\(y-y_{1}=m\left ( x-x_{1} \right )\)
\(y-\left ( -3 \right )=\frac{9}{5}\left ( x-6 \right )\)
Simplify the equation:
\(y+3=\frac{9}{5}x-\frac{54}{5}\)
\(y=\frac{9}{5}x-\frac{54}{5}-\frac{15}{3}\)
Combine the constants:
\(y=\frac{9}{5}x-\frac{69}{5}\)
So, the equation of line L in the form y=mx+c is
\(y=\frac{9}{5}x-\frac{69}{5}\)
Question 2
Topic: Topic: SL 2.8
Given
- Function f as f(x) = \(\frac{7x + 7}{2x – 4}\) for \(x\epsilon \mathbb{R}\), x ≠ 2.
Find
- root of the function f(x).
- for graph of function f(x)
- equation of vertical asymptote
- equation of horizontal asymptote
- \(f^{-1}\) (x)
▶️Answer/Explanation
Answer:
(a) recognizing f(x) = 0
x = -1
Roots of the Function (where ):
Set the numerator equal to zero and solve for
:
7x + 7 = 0
7x = -7
x = -1
Therefore, the root is x = -1
(b) (i) x = 2 (must be an equation with x)
(ii) y = \(\frac{7}{2}\) (must be an equation with y)
(i) Equation of Vertical Asymptote (where approaches infinity or negative infinity):
The vertical asymptote occurs where the denominator is equal to zero (excluding any values of
that make the numerator zero since those would be roots). In this case:
Therefore, the equation of the vertical asymptote is
.
(ii)Equation of Horizontal Asymptote (where approaches a constant as
approaches positive or negative infinity):
To find the horizontal asymptote, look at the degrees of the numerator and denominator. In this case, the degrees are the same (both 1). Therefore, you can find the horizontal asymptote by dividing the leading coefficient of the numerator by the leading coefficient of the denominator:
\(\lim_{x\rightarrow \infty }=\frac{7x+4}{2x-4}=\frac{7}{2}=3.5\)
So, the equation of the horizontal asymptote is \(\frac{7}{2}=3.5\)
(c) EITHER
interchanging x and y
2xy – 4x = 7y + 7
correct working with y terms on the same side: 2xy – 7y = 4x + 7
OR
2yx – 4y = 7x +7
correct working with x terms on the same side: 2yx – 7x = 4x + 7
interchanging x and y OR making x the subject x = \(\frac{4y + 7}{2y – 7}\)
THEN
\(f^{-1}\)(x) = \(\frac{4x + 7}{2x – 7}\) (or equivalent) (\(x ≠ \frac{7}{2}\))
Start with the original function:
\(y=\frac{7x+4}{2x-4}\)
Swap
and
:
\(x=\frac{7y+4}{2y-4}\)
Solve for
:
- Cross-multiply to get rid of the fraction:
- Move the
terms to one side and the constant terms to the other: - Factor out
on the left side: - Solve for
:\(y=\frac{4x+7}{2x-7}\)
So, the inverse function is \(f^{-1}\left ( x \right )=\frac{4x+7}{2x-7}\)