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Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If l

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New post 04 Oct 2018, 00:58
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[Math Revolution GMAT math practice question]

Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If lines AB and BC are perpendicular to each other, p=?

A. -2
B. -1
C. 0
D. 1
E. 2

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Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If l  [#permalink]

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New post 04 Oct 2018, 01:20
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If lines AB and BC are perpendicular to each other, p=?

A. -2
B. -1
C. 0
D. 1
E. 2


Perpendicular bisectors have a negative reciprocal of their slopes.
A (1,4)
B (2,2)
C (p,1)

Slope of AB= 4-2/1-2=-2
Slope of BC will be 1/2

or slope of BC= 2-1/2-p=1/2---- cross multiply
(2-1)2=2-p
4-2=2-p
4-2-2=p
0=p

Therefore p=0
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Re: Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If l  [#permalink]

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New post 04 Oct 2018, 07:05
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If lines AB and BC are perpendicular to each other, p=?

A. -2
B. -1
C. 0
D. 1
E. 2

Line AB is oblique (non-horizontal and non-vertical) therefore line BC is also oblique, and the product of their slopes must be -1.

\(?\,\, = \,p\)

\({\text{slope}}{\,_{\overleftrightarrow {{\text{AB}}}}}\,\, = \,\,\,\frac{{4 - 2}}{{1 - 2}}\,\, = \, - 2\,\)

\(\frac{1}{2}\,\,{\text{ = }}\,\,{\text{slope}}{\,_{\overleftrightarrow {{\text{BC}}}}}\,\,\,\,{\text{ = }}\,\,\frac{{2 - 1}}{{2 - p}}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,p = 0\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If l  [#permalink]

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New post 07 Oct 2018, 18:37
=>

The slope of the line segment joining the two points (x1,y1) and (x2,y2) is (y1-y2)/(x1-x2). The product of the slopes of perpendicular lines is \(-1\).

The slope of the line AB is \(\frac{(4-2)}{(1-2)} = -2.\)
Since the two lines AB and BC are perpendicular each other, the slope of BC, which is \(\frac{(2-1)}{(2-p)}\), must be \(\frac{1}{2}.\)
So,
\(\frac{(2-1)}{(2-p)}=\frac{1}{2}\)
\(\frac{1}{(2-p)}=\frac{1}{2}.\)
Thus, \(2 – p = 2\), and \(p = 0\).

Therefore, the answer is C.
Answer: C
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Re: Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If l   [#permalink] 07 Oct 2018, 18:37
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