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In the figure above
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04 May 2018, 16:59
04 May 2018, 16:59
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Question Stats:
42% (00:47) correct
57% (02:01) wrong based on 7 sessions
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In the figure above, what are the values of X and Y?
A. \(x= \frac{ak}{b}; y = b - k\)
B. \(x = a; y= b\)
C. \(x = \frac{a}{b}; y = b - k\)
D. \(x = \frac{a}{bk}; y=b\)
E. \(x = \frac{ak}{b}; y = b + k\)
Re: In the figure above
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05 May 2018, 11:04
05 May 2018, 11:04
Anyone have an answer to this question? and possibly an explanation too please?
Re: In the figure above
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05 May 2018, 12:24
05 May 2018, 12:24
2
Expert Reply
These types of questions are prime candidate for elemination methods.
From the figure we can see that the point \((x,y)\) lies on the line which starts from (a,0) and ends at (0,b). The y coordinate of the point (x,y) is shown to be k units below point (0,b).
So \(y = k-b\).
Eleminate all other options apart from A and C.
Now if you cant intuitively figure out that option A is most likely. Just write the equation of the line.
\(y= mx + c\) where m is the slope and c is the intercept. So points (a,0) and (0, b) must both satisfy the equation.
So substitute (a,0) \(0= m \times a + c\) and substitute (0,b) \(b= m \times 0 + c\). Solve for m and c
\(c=b\) and \(m= \frac{-b}{a}\).
So equation of line is \(y=\frac{-b}{a} \times x + b\).
Put value of y as b-k to solve for x \(b-k=\frac{-b}{a} \times x +b\) so \(x=\frac{ak}{b}\).
Hence option A.
Note there is an easier form of the equation of line if the line if the points of intersection with x and y axis are known. That form is
\(\frac{x}{a}+ \frac{y}{b}=1\).
From the figure we can see that the point \((x,y)\) lies on the line which starts from (a,0) and ends at (0,b). The y coordinate of the point (x,y) is shown to be k units below point (0,b).
So \(y = k-b\).
Eleminate all other options apart from A and C.
Now if you cant intuitively figure out that option A is most likely. Just write the equation of the line.
\(y= mx + c\) where m is the slope and c is the intercept. So points (a,0) and (0, b) must both satisfy the equation.
So substitute (a,0) \(0= m \times a + c\) and substitute (0,b) \(b= m \times 0 + c\). Solve for m and c
\(c=b\) and \(m= \frac{-b}{a}\).
So equation of line is \(y=\frac{-b}{a} \times x + b\).
Put value of y as b-k to solve for x \(b-k=\frac{-b}{a} \times x +b\) so \(x=\frac{ak}{b}\).
Hence option A.
Note there is an easier form of the equation of line if the line if the points of intersection with x and y axis are known. That form is
\(\frac{x}{a}+ \frac{y}{b}=1\).
