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01 Aug 2021, 03:28
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47% (02:03) correct
53%
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In the right coordinate plane, is the distance between the point (a, b) and the point (c, d) greater than 3?
1) a = c + 1 and b = d + 7
2) a = d + 4 and c = b − 4
1) a = c + 1 and b = d + 7
2) a = d + 4 and c = b − 4
02 Aug 2021, 01:04
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sjuniv32
The distance formula is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
1) a = c + 1 and b = d + 7
Now b and d are y co-ordinates and are 7 units away, so the points have to be greater than 7 units apart. SUFFICIENT
OR
\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)=\(\sqrt{(c-a)^2+(d-b)^2}\)
Substitute the values of a and b
\(\sqrt{(c-(c+1))^2+(d-(d+7))^2}=\sqrt{(-1)^2+(-7)^2}=\sqrt{50}\), which is greater than 3.
Sufficient
2) a = d + 4 and c = b − 4
\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)=\(\sqrt{(c-a)^2+(d-b)^2}\)
Substitute the values of a=d+4 and b=c+4
\(\sqrt{(c-(d+4))^2+(d-(c+4))^2}=\sqrt{((c-d)-4)^2+((d-c)-4)^2}=\sqrt{(c-d)^2-8*(c-d)+16+(d-c)^2-8(d-c)+16}\)
\(\sqrt{2*(c-d)^2-8*(c-d)+8(c-d)+2*16}=\sqrt{32+2*(c-d)^2}\)
So minimum value of \(\sqrt{32+2*(c-d)^2}\) is \(\sqrt{32}\) when c=d.
Thus distance is greater than 3.
Sufficient
D
30 Aug 2022, 16:50
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