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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Points A(1,4), B(2,2) and C(p,1) lie on the x-y coordinate plane. If l

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8261
GMAT 1: 760 Q51 V42 GPA: 3.82
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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Difficulty:   45% (medium)

Question Stats: 64% (01:53) correct 36% (02:04) wrong based on 65 sessions

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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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Manager  S
Joined: 17 Mar 2018
Posts: 71
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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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
GMATH Teacher P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
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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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.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8261
GMAT 1: 760 Q51 V42 GPA: 3.82
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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=>

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 $$p = 0$$.

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